This is “Exponential and Logarithmic Functions”, chapter 7 ...€¦ · Chapter 7 Exponential and Logarithmic Functions 7.1 Composition and Inverse Functions 1578 The horizontal

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Chapter 7

Exponential and Logarithmic Functions

1568

7.1 Composition and Inverse Functions

LEARNING OBJECTIVES

1. Perform function composition.2. Determine whether or not given functions are inverses.3. Use the horizontal line test.4. Find the inverse of a one-to-one function algebraically.

Composition of Functions

In mathematics, it is often the case that the result of one function is evaluated byapplying a second function. For example, consider the functions defined byf (x) = x2 and g (x) = 2x + 5. First, g is evaluated where x = −1 and then theresult is squared using the second function, f.

This sequential calculation results in 9. We can streamline this process by creating anew function defined by f (g (x)), which is explicitly obtained by substituting g (x)into f (x) .

Therefore, f (g (x)) = 4x2 + 20x + 25and we can verify that when x = −1 theresult is 9.

f (g (x))==

=

f (2x + 5)(2x + 5)24x2 + 20x + 25

Chapter 7 Exponential and Logarithmic Functions

1569

The calculation above describes composition of functions1, which is indicatedusing the composition operator2 (○). If given functions f and g,

The notation f○g is read, “f composed with g.” This operation is only defined forvalues, x, in the domain of g such that g (x) is in the domain of f.

f (g (−1))===

4(−1)2 + 20 (−1) + 254 − 20 + 259

(f○g) (x) = f (g (x)) Composition of Functions

1. Applying a function to theresults of another function.

2. The open dot used to indicatethe function composition

(f○g) (x) = f (g (x)) .


7.1 Composition and Inverse Functions 1570

Example 1

Given f (x) = x2 − x + 3 and g (x) = 2x − 1 calculate:

a. (f○g) (x) .b. (g○f ) (x) .

Solution:

a. Substitute g into f.

b. Substitute f into g.

Answer:

a. (f○g) (x) = 4x2 − 6x + 5

(f○g) (x)=====

f (g (x))f (2x − 1)(2x − 1)2 − (2x − 1) + 34x2 − 4x + 1 − 2x + 1 + 34x2 − 6x + 5

(g○f ) (x)==

=

==

g (f (x))g (x2 − x + 3)2 (x2 − x + 3) − 1

2x2 − 2x + 6 − 12x2 − 2x + 5



b. (g○f ) (x) = 2x2 − 2x + 5

The previous example shows that composition of functions is not necessarilycommutative.



Example 2

Given f (x) = x3 + 1 and g (x) = 3x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3 find (f○g) (4) .Solution:

Begin by finding (f○g) (x) .

Next, substitute 4 in for x.

Answer: (f○g) (4) = 12

Functions can be composed with themselves.

(f○g) (x)==

=

==

f (g (x))f ( 3x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3 )( 3x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯√3 )

3+ 1

3x − 1 + 13x

(f○g) (x)(f○g) (4)

=

==

3x

3 (4)12



Example 3

Given f (x) = x2 − 2 find (f○f ) (x) .Solution:

Answer: (f○f ) (x) = x4 − 4x2 + 2

Try this! Given f (x) = 2x + 3 and g (x) = x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯

√ find (f○g) (5) .Answer: 7

(click to see video)

Inverse Functions

Consider the function that converts degrees Fahrenheit to degrees Celsius:

C (x) = 59 (x − 32) .We can use this function to convert 77°F to degrees Celsius as

follows.

(f○f ) (x)==

=

==

f (f (x))f (x2 − 2)(x2 − 2)2 − 2

x4 − 4x2 + 4 − 2x4 − 4x2 + 2



http://www.youtube.com/v/cwEGsJa5Tlc

Therefore, 77°F is equivalent to 25°C. If we wish to convert 25°C back to degrees

Fahrenheit we would use the formula: F (x) = 95 x + 32.

Notice that the two functions C and F each reverse the effect of the other.

This describes an inverse relationship. In general, f and g are inverse functions if,

In this example,

C (77)=

=

=

59

(77 − 32)

59 (45)25

F (25)===

95 (25) + 32

45 + 3277

(f○g) (x)(g○f ) (x)

=

=

f (g (x)) = x f or all x in the domain of g and

g (f (x)) = x f or all x in the domain of f .



Example 4

Verify algebraically that the functions defined by f (x) = 12 x − 5and

g (x) = 2x + 10 are inverses.

Solution:

Compose the functions both ways and verify that the result is x.

(f○g) (x)===

==

f (g (x))f (2x + 10)12 (2x + 10) − 5

x + 5 − 5x ✓

(g○f ) (x)==

=

==

g (f (x))g ( 12 x − 5)2 ( 12 x − 5) + 10

x − 10 + 10x ✓

Answer: Both (f○g) (x) = (g○f ) (x) = x; therefore, they are inverses.

Next we explore the geometry associated with inverse functions. The graphs of bothfunctions in the previous example are provided on the same set of axes below.

C (F (25))F (C (77))

=

=

C (77) = 25

F (25) = 77



Note that there is symmetry about the line y = x ; the graphs of f and g are mirrorimages about this line. Also notice that the point (20, 5) is on the graph of f and that(5, 20) is on the graph of g. Both of these observations are true in general and wehave the following properties of inverse functions:

1. The graphs of inverse functions are symmetric about the line y = x.2. If (a, b)is on the graph of a function, then (b, a)is on the graph of its

inverse.

Furthermore, if g is the inverse of f we use the notation g = f −1 .Here f −1 is read,“f inverse,” and should not be confused with negative exponents. In other words,

f −1 (x) ≠ 1f (x) and we have,

(f○f −1) (x)(f −1○f) (x)

=

=

f (f −1 (x)) = x

f −1 (f (x)) = x

and



Example 5

Verify algebraically that the functions defined by f (x) = 1x − 2and

f −1 (x) = 1x+2 are inverses.

Solution:

Compose the functions both ways to verify that the result is x.

(f○f −1) (x)==

=

=

==

f (f −1 (x))f ( 1

x+2 )1

( 1x+2 )

− 2

x+21 − 2

x + 2 − 2x ✓

(f −1○f) (x)==

=

=

=

f −1 (f (x))f −1 ( 1x − 2)

1

( 1x −2)+2

11x

x ✓

Answer: Since (f○f −1) (x) = (f −1○f) (x) = xthey are inverses.

Recall that a function is a relation where each element in the domain correspondsto exactly one element in the range. We use the vertical line test to determine if agraph represents a function or not. Functions can be further classified using aninverse relationship. One-to-one functions3 are functions where each value in therange corresponds to exactly one element in the domain. The horizontal line test4

is used to determine whether or not a graph represents a one-to-one function. If ahorizontal line intersects a graph more than once, then it does not represent a one-to-one function.

3. Functions where each value inthe range corresponds toexactly one value in thedomain.

4. If a horizontal line intersectsthe graph of a function morethan once, then it is not one-to-one.



The horizontal line represents a value in the range and the number of intersectionswith the graph represents the number of values it corresponds to in the domain.The function defined by f (x) = x3 is one-to-one and the function defined byf (x) = |x| is not. Determining whether or not a function is one-to-one isimportant because a function has an inverse if and only if it is one-to-one. In otherwords, a function has an inverse if it passes the horizontal line test.

Note: In this text, when we say “a function has an inverse,” we mean that there is

another function, f −1 , such that (f○f −1) (x) = (f −1○f) (x) = x.



Example 6

Determine whether or not the given function is one-to-one.

Solution:

Answer: The given function passes the horizontal line test and thus is one-to-one.

In fact, any linear function of the form f (x) = mx + b where m ≠ 0, is one-to-oneand thus has an inverse. The steps for finding the inverse of a one-to-one functionare outlined in the following example.



Example 7

Find the inverse of the function defined by f (x) = 32 x − 5.

Solution:

Before beginning this process, you should verify that the function is one-to-one. In this case, we have a linear function where m ≠ 0 and thus it is one-to-one.

• Step 1: Replace the function notation f (x) with y.

• Step 2: Interchange x and y. We use the fact that if (x, y) is a point

on the graph of a function, then (y, x) is a point on the graph ofits inverse.

• Step 3: Solve for y.

f (x)

y

=

=

32x − 5

32x − 5

x =32y − 5



• Step 4: The resulting function is the inverse of f. Replace y withf −1 (x) .

• Step 5: Check.

(f○f −1) (x)= f (f −1 (x))= f ( 23 x + 10

3 )= 3

2 ( 23 x + 103 ) − 5

= x + 5 − 5= x ✓

(f −1○f) (x)= f −1 (f (x))= f −1 ( 32 x − 5)= 2

3 ( 32 x − 5) + 103

= x − 103 + 10

3

= x ✓

Answer: f −1 (x) = 23 x + 10

3

If a function is not one-to-one, it is often the case that we can restrict the domain insuch a way that the resulting graph is one-to-one. For example, consider thesquaring function shifted up one unit, g (x) = x2 + 1.Note that it does not passthe horizontal line test and thus is not one-to-one. However, if we restrict the

x

x + 5

23

⋅ (x + 5)23x +

103

=

=

=

=

32y − 5

32y

23

⋅32y

y

f −1 (x) =23x +

103



domain to nonnegative values, x ≥ 0, then the graph does pass the horizontal linetest.

On the restricted domain, g is one-to-one and we can find its inverse.



Example 8

Find the inverse of the function defined by g (x) = x2 + 1 where x ≥ 0.

Solution:

Begin by replacing the function notation g (x) with y.

Interchange x and y.

Solve for y.

Since y ≥ 0 we only consider the positive result.

g (x)y

==x2 + 1x2 + 1 where x ≥ 0

x = y2 + 1 where y ≥ 0

x

x − 1± x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯√

===

y2 + 1y2

y



Answer: g−1 (x) = x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯

√ . The check is left to the reader.

The graphs in the previous example are shown on the same set of axes below. Takenote of the symmetry about the line y = x.

y

g−1 (x)

=

=

x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯√

x − 1⎯ ⎯⎯⎯⎯⎯⎯⎯√



Example 9

Find the inverse of the function defined by f (x) = 2x+1x−3 .

Solution:

Use a graphing utility to verify that this function is one-to-one. Begin byreplacing the function notation f (x) with y.

Interchange x and y.

Solve for y.

f (x)

y

=

=

2x + 1x − 32x + 1x − 3

x =2y + 1y − 3

x

x (y − 3)xy − 3x

=

==

2y + 1y − 32y + 12y + 1



Obtain all terms with the variable y on one side of the equation and everythingelse on the other. This will enable us to treat y as a GCF.

Answer: f −1 (x) = 3x+1x−2 .The check is left to the reader.

Try this! Find the inverse of f (x) = x + 1⎯ ⎯⎯⎯⎯⎯⎯⎯√3 − 3.

Answer: f −1 (x) = (x + 3)3 − 1


xy − 3xxy − 2yy (x − 2)

y

===

=

2y + 13x + 13x + 13x + 1x − 2



http://www.youtube.com/v/MwSB9WuYfCA

KEY TAKEAWAYS

• The composition operator (○) indicates that we should substitute one

function into another. In other words, (f○g) (x) = f (g (x))indicates that we substitute g (x) into f (x) .

• If two functions are inverses, then each will reverse the effect of the

other. Using notation, (f○g) (x) = f (g (x)) = x and

(g○f ) (x) = g (f (x)) = x.• Inverse functions have special notation. If g is the inverse of f , then we

can write g (x) = f −1 (x) . This notation is often confused withnegative exponents and does not equal one divided by f (x) .

• The graphs of inverses are symmetric about the line y = x. If (a, b) is

a point on the graph of a function, then (b, a) is a point on the graph

of its inverse.• If each point in the range of a function corresponds to exactly one value

in the domain then the function is one-to-one. Use the horizontal linetest to determine whether or not a function is one-to-one.

• A one-to-one function has an inverse, which can often be found byinterchanging x and y, and solving for y. This new function is the inverseof the original function.



TOPIC EXERCISES

PART A : COMPOSITION OF FUNCTIONS

Given the functions defined by f and g find (f○g) (x) and

(g○f ) (x) .1. f (x) = 4x − 1 , g (x) = 3x2. f (x) = −2x + 5 , g (x) = 2x3. f (x) = 3x − 5 , g (x) = x − 44. f (x) = 5x + 1 , g (x) = 2x − 3

5. f (x) = x 2 − x + 1, g (x) = 2x − 1

6. f (x) = x 2 − 3x − 2 , g (x) = x − 2

7. f (x) = x 2 + 3, g (x) = x 2 − 5

8. f (x) = 2x 2 , g (x) = x 2 − x

9. f (x) = 8x 3 + 5, g (x) = x − 5⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3

10. f (x) = 27x 3 − 1 , g (x) = x + 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3

11. f (x) = 1x+5 , g (x) = 1

x

12. f (x) = 1x − 3, g (x) = 3

x+3

13. f (x) = 5 x⎯⎯

√ , g (x) = 3x − 2

14. f (x) = 2x⎯ ⎯⎯⎯

√ , g (x) = 4x + 1

15. f (x) = 12x , g (x) = x 2 + 8

16. f (x) = 2x − 1 , g (x) = 1x+1

17. f (x) = 1−x2x , g (x) = 1

2x+1

18. f (x) = 2xx+1 , g (x) = x+1

x



Given the functions defined by f (x) = 3x 2 − 2, g (x) = 5x + 1 ,

and h (x) = x⎯⎯

√ , calculate the following.

19. (f○g) (2)20. (g○f ) (−1)

21. (g○f ) (0)22. (f○g) (0)23. (f○h) (3)24. (g○h) (16)25. (h○g) ( 35 )26. (h○f ) (−3)

Given the functions defined by f (x) = x + 3⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3 , g (x) = 8x 3 − 3,and h (x) = 2x − 1 , calculate the following.

27. (f○g) (1)28. (g○f ) (−2)

29. (g○f ) (0)30. (f○g) (−2)

31. (f○h) (−1)

32.

33. (h○f ) (24)34. (g○h) (0)

Given the function, determine (f○f ) (x) .35. f (x) = 3x − 1

(h○g) (−12 )



36. f (x) = 25 x + 1

37. f (x) = x 2 + 5

38. f (x) = x 2 − x + 6

39. f (x) = x 3 + 2

40. f (x) = x 3 − x

41. f (x) = 1x+1

42. f (x) = x+12x

PART B : INVERSE FUNCTIONS

Are the given functions one-to-one? Explain.

43.



44.

45.

46.

47. f (x) = x + 1

48. g (x) = x 2 + 1



49. h (x) = |x| + 1

50. r (x) = x 3 + 1

51. f (x) = x + 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√

52. g (x) = 3

Given the graph of a one-to-one function, graph its inverse.

53.

54.



55.

56.



57.

58.

Verify algebraically that the two given functions are inverses. In other

words, show that (f○f −1) (x) = x and (f −1○f) (x) = x.



59. f (x) = 3x − 4 , f −1 (x) = x+43

60. f (x) = −5x + 1 , f −1 (x) = 1−x5

61. f (x) = − 23 x + 1, f −1 (x) = − 3

2 x + 32

62. f (x) = 4x − 13 , f −1 (x) = 1

4 x + 112

63. f (x) = x − 8⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√ , f −1 (x) = x 2 + 8, x ≥ 0

64. f (x) = 6x⎯ ⎯⎯⎯

√3 − 3, f −1 (x) = (x+3)3

6

65. f (x) = xx+1 , f −1 (x) = x

1−x

66. f (x) = x−33x , f −1 (x) = 3

1−3x

67. f (x) = 2(x − 1)3 + 3, f −1 (x) = 1 + x−32

⎯ ⎯⎯⎯⎯⎯√368. f (x) = 5x − 1

⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3 + 4, f −1 (x) = (x−4)3+1

5

Find the inverses of the following functions.

69. f (x) = 5x

70. f (x) = 12 x

71. f (x) = 2x + 572. f (x) = −4x + 3

73. f (x) = − 23 x + 1

3

74. f (x) = − 12 x + 3

4

75. g (x) = x 2 + 5, x ≥ 0

76. g (x) = x 2 − 7, x ≥ 0

77. f (x) = (x − 5) 2 , x ≥ 5

78. f (x) = (x + 1)2 , x ≥ −1

79. h (x) = 3x 3 + 5



80. h (x) = 2x 3 − 1

81. f (x) = (2x − 3) 3

82. f (x) = (x + 4)3 − 1

83.

84.

85.

86.

87.

88.

89.

90.

91. g (x) = 5x + 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3

92. g (x) = 4x − 3⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3

93. f (x) = x − 6⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3 − 4

94. f (x) = 2 x + 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3 + 5

95. h (x) = x + 1⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√5 − 3

96. h (x) = x − 8⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√5 + 1

97. f (x) = mx + b , m ≠ 0

98. f (x) = ax 2 + c, x ≥ 0

99. f (x) = ax 3 + d

100. f (x) = a(x − h) 2 + k, x ≥ h

g (x) =2

x 3 + 1g (x) =

1x 3 − 2

f (x) =5

x + 1f (x) =

12x − 9

f (x) =x + 5x − 5

f (x) =3x − 42x − 1

h (x) =x − 510x

h (x) =9x + 13x



Graph the function and its inverse on the same set of axes.

101. f (x) = x + 2

102. f (x) = 23 x − 4

103. f (x) = −2x + 2

104. f (x) = − 13 x + 4

105. g (x) = x 2 − 2, x ≥ 0

106. g (x) = (x − 2)2 , x ≥ 2

107. h (x) = x 3 + 1

108. h (x) = (x + 2)3 − 2

109. f (x) = 2 − x⎯⎯

√

110. f (x) = −x⎯ ⎯⎯⎯⎯√ + 1

PART C : D ISCUSSION BOARD

111. Is composition of functions associative? Explain.

112. Explain why C (x) = 59 (x − 32) and F (x) = 9

5 x + 32 define inverse

functions. Prove it algebraically.

113. Do the graphs of all straight lines represent one-to-one functions? Explain.

114. If the graphs of inverse functions intersect, then how can we find the point ofintersection? Explain.



ANSWERS

1. (f○g) (x) = 12x − 1 ; (g○f ) (x) = 12x − 3

3. (f○g) (x) = 3x − 17 ; (g○f ) (x) = 3x − 9

5. (f○g) (x) = 4x 2 − 6x + 3 ; (g○f ) (x) = 2x 2 − 2x + 1

7. (f○g) (x) = x 4 − 10x 2 + 28 ; (g○f ) (x) = x 4 + 6x 2 + 4

9. (f○g) (x) = 8x − 35 ; (g○f ) (x) = 2x

11. (f○g) (x) = x5x+1 ; (g○f ) (x) = x + 5

13. (f○g) (x) = 5 3x − 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√ ; (g○f ) (x) = 15 x⎯⎯

√ − 2

15.

17. (f○g) (x) = x ; (g○f ) (x) = x

19. 361

21. −9

23. 7

25. 2

27. 2

29. 21

31. 0

33. 5

35. (f○f ) (x) = 9x − 4

37. (f○f ) (x) = x 4 + 10x 2 + 30

39. (f○f ) (x) = x 9 + 6x 6 + 12x 3 + 10

41. (f○f ) (x) = x+1x+2

(f○g) (x) = 12x 2 + 16

;

(g○f ) (x) = 1 + 32x 2

4x 2



43. No, fails the HLT

45. Yes, passes the HLT

47. Yes, its graph passes the HLT.

49. No, its graph fails the HLT.

51. Yes, its graph passes the HLT.

53.

55.



57.

59. Proof

61. Proof

63. Proof

65. Proof

67. Proof

69. f −1 (x) = x5

71.

73.

75. g−1 (x) = x − 5⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√

77. f −1 (x) = x⎯⎯

√ + 5

79.

81.

f −1 (x) =12x −

52

f −1 (x) = −32x +

12

h−1 (x) =x − 53

⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3

f −1 (x) =x⎯⎯

√3 + 32



83.

85.

87.

89.

91.

93. f −1 (x) = (x + 4)3 + 6

95. h−1 (x) = (x + 3)5 − 1

97.

99.

101.

g−1 (x) =2 − x

x

⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3f −1 (x) =

5 − x

x

f −1 (x) =5 (x + 1)x − 1

h−1 (x) = −5

10x − 1g−1 (x) =

x 3 − 25

f −1 (x) =x − b

m

f −1 (x) =x − d

a

⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3



103.

105.



107.

109.

111. Answer may vary




7.2 Exponential Functions and Their Graphs

LEARNING OBJECTIVES

1. Identify and evaluate exponential functions.2. Sketch the graph of exponential functions and determine the domain

and range.3. Identify and graph the natural exponential function.4. Apply the formulas for compound interest.

Exponential Functions

At this point in our study of algebra we begin to look at transcendental functions orfunctions that seem to “transcend” algebra. We have studied functions withvariable bases and constant exponents such as x2 or y−3 . In this section we explorefunctions with a constant base and variable exponents. Given a real number b > 0where b ≠ 1 an exponential function5 has the form,

For example, if the base b is equal to 2, then we have the exponential functiondefined by f (x) = 2x . Here we can see the exponent is the variable. Up to thispoint, rational exponents have been defined but irrational exponents have not.

Consider 2 7√ , where the exponent is an irrational number in the range,

We can use these bounds to estimate 2 7√ ,

f (x) = bx Exponential Function

2.64 < 7⎯⎯√ < 2.65

5. Any function with a definitionof the form f (x) = bx whereb > 0 and b ≠ 1.


1605

Using rational exponents in this manner, an approximation of 2 7√ can be obtainedto any level of accuracy. On a calculator,

Therefore the domain of any exponential function consists of all real numbers

(−∞, ∞) . Choose some values for x and then determine the corresponding y-values.

Because exponents are defined for any real number we can sketch the graph using acontinuous curve through these given points:

22.64 < 2 7√ < 22.65

6.23 < 2 7√ < 6.28

2 ^ 7⎯⎯√ ≈ 6.26

x

−2

−1

0

1

2

7⎯⎯√

y

1412

1

2

4

6.26

f (x) = 2x

y = 2−2 =122

=14

y = 2−1 =121

=12

y = 20 = 1

y = 21 = 2

y = 22 = 4

y = 2 7√ ≈ 6.26

Solutions

(−2,14)

(−1,12)

(0, 1)(1, 2)(2, 4)(2.65, 6.26)


7.2 Exponential Functions and Their Graphs 1606

It is important to point out that as x approaches negative infinity, the resultsbecome very small but never actually attain zero. For example,

This describes a horizontal asymptote at y = 0, the x-axis, and defines a lowerbound for the range of the function: (0, ∞) .The base b of an exponential function affects the rate at which it grows. Below wehave graphed y = 2x , y = 3x , and y = 10x on the same set of axes.

f (−5)f (−10)

f (−15)

=

=

=

2−5 =125

≈ 0.03125

2−10 =1210

≈ 0.0009766

2−15 =1

2−15≈ .00003052



Note that all of these exponential functions have the same y-intercept, namely(0, 1) .This is because f (0) = b0 = 1 for any function defined using the formf (x) = bx . As the functions are read from left to right, they are interpreted asincreasing or growing exponentially. Furthermore, any exponential function of thisform will have a domain that consists of all real numbers (−∞, ∞) and a range that

consists of positive values (0, ∞) bounded by a horizontal asymptote at y = 0.



Example 1

Sketch the graph and determine the domain and range: f (x) = 10x + 5.

Solution:

The base 10 is used often, most notably with scientific notation. Hence, 10 iscalled the common base. In fact, the exponential function y = 10x is so

important that you will find a button 10x dedicated to it on most modern

scientific calculators. In this example, we will sketch the basic graph y = 10xand then shift it up 5 units.

Note that the horizontal asymptote of the basic graph y = 10x was shifted up 5units to y = 5 (shown dashed). Take a minute to evaluate a few values of x withyour calculator and convince yourself that the result will never be less than 5.

Answer:

Domain: (−∞, ∞) ; Range: (5, ∞)



Next consider exponential functions with fractional bases 0 < b < 1. For example,

f (x) = ( 12 )x is an exponential function with base b = 12 .

Plotting points we have,

Reading the graph from left to right, it is interpreted as decreasing exponentially.The base affects the rate at which the exponential function decreases or decays.

Below we have graphed y = ( 12 )x , y = ( 13 )x , and y = ( 110 )x on the same set of

axes.

x

−2

−1

0

1

2

y

4

2

1

12

14

f (x) = ( 12)

x

f ( 12) = ( 1

2)−2

=1−2

2−2=

22

12= 4

f ( 12) = ( 1

2)−1

=1−1

2−1=

21

11= 2

f ( 12) = ( 1

2)0

= 1

f ( 12) = ( 1

2)1

=12

f ( 12) = ( 1

2)2

=14

Solutions

(−2, 4)

(−1, 2)

(0, 1)

(1,12)

(2,14)



Recall that x−1 = 1x and so we can express exponential functions with fractional

bases using negative exponents. For example,

Furthermore, given that f (x) = 2x we can see g (x) = f (−x) = 2−x and canconsider g to be a reflection of f about the y-axis.

In summary, given b > 0

g (x) = ( 12 )

x

=1x

2x=

12x

= 2−x .



And for both cases,

Furthermore, note that the graphs pass the horizontal line test and thusexponential functions are one-to-one. We use these basic graphs, along with thetransformations, to sketch the graphs of exponential functions.

Domain :

Range :y − intercept :Asymptote :

(−∞, ∞)(0, ∞)(0, 1)y = 0



Example 2

Sketch the graph and determine the domain and range: f (x) = 5−x − 10.

Solution:

Begin with the basic graph y = 5−x and shift it down 10 units.

The y-intercept is (0, −9) and the horizontal asymptote is y = −10.

Answer:

Domain: (−∞, ∞); Range: (−10, ∞)

Note: Finding the x-intercept of the graph in the previous example is left for a latersection in this chapter. For now, we are more concerned with the general shape ofexponential functions.



Example 3

Sketch the graph and determine the domain and range: g (x) = −2x−3 .

Solution:

Begin with the basic graph y = 2x and identify the transformations.

Note that the horizontal asymptote remains the same for all of thetransformations. To finish we usually want to include the y-intercept.Remember that to find the y-intercept we set x = 0.

y

y

y

===

2x

−2x

−2x−3

Basic graph

Ref lection about the x-axis

Shif t right 3 units

g (0)==

=

=

−20−3

−2−3

−123

−18



Therefore the y-intercept is (0, − 18 ).

Answer:

Domain: (−∞, ∞) ; Range: (−∞, 0)

Try this! Sketch the graph and determine the domain and range:f (x) = 2x−1 + 3.

Answer:

Domain: (−∞, ∞) ; Range: (3, ∞)(click to see video)



http://www.youtube.com/v/WDcObqDtE7M

Natural Base e

Some numbers occur often in common applications. One such familiar number is pi(π), which we know occurs when working with circles. This irrational number has a

dedicated button on most calculators π and approximated to five decimal places,

π ≈ 3.14159.Another important number e occurs when working with exponentialgrowth and decay models. It is an irrational number and approximated to fivedecimal places, e ≈ 2.71828.This constant occurs naturally in many real-worldapplications and thus is called the natural base. Sometimes e is called Euler’sconstant in honor of Leonhard Euler (pronounced “Oiler”).

Figure 7.1

Leonhard Euler (1707–1783)

In fact, the natural exponential function,

is so important that you will find a button ex dedicated to it on any modern

scientific calculator. In this section, we are interested in evaluating the naturalexponential function for given real numbers and sketching its graph. To evaluatethe natural exponential function, defined by f (x) = ex where x = −2 using acalculator, you may need to apply the shift button. On many scientific calculatorsthe caret will display as follows,

f (x) = ex

f (−2) = e ^ (−2) ≈ 0.13534



After learning how to use your particular calculator, you can now sketch the graphby plotting points. (Round off to the nearest hundredth.)

Plot the points and sketch the graph.

Note that the function is similar to the graph of y = 3x . The domain consists of allreal numbers and the range consists of all positive real numbers. There is anasymptote at y = 0 and a y-intercept at (0, 1) .We can use the transformations tosketch the graph of more complicated exponential functions.

x

−2−1012

y

0.140.3712.727.39

f (x) = ex

f (−2) = e−2 = 0.14f (−1) = e−1 = 0.37f (0) = e0 = 1f (1) = e1 = 2.72f (2) = e2 = 7.39

Solutions

(−2, 0.14)(−1, 0.37)(0, 1)(1, 2.72)(2, 7.39)



Example 4

Sketch the graph and determine the domain and range: g (x) = ex+2 − 3.

Solution:

Identify the basic transformations.

To determine the y-intercept set x = 0.

Therefore the y-intercept is (0, e2 − 3) .

y

y

y

===

ex

ex+2

ex+2 − 3

Basic graph

Shif t lef t 2 unitsShif t down 3 units

g (0)==≈

e0+2 − 3e2 − 34.39



Answer:

Domain: (−∞, ∞) ; Range: (−3, ∞)

Try this! Sketch the graph and determine the domain and range:f (x) = e−x + 2.

Answer:

Domain: (−∞, ∞) ; Range: (2, ∞)(click to see video)

Compound Interest Formulas

Exponential functions appear in formulas used to calculate interest earned in mostregular savings accounts. Compound interest occurs when interest accumulated for



http://www.youtube.com/v/FkwBpg2Xpb8

one period is added to the principal investment before calculating interest for thenext period. The amount accrued in this manner over time is modeled by thecompound interest formula6:

Here the amount A depends on the time t in years the principal P is accumulatingcompound interest at an annual interest rate r. The value n represents the numberof times the interest is compounded in a year.

A (t) = P(1 +r

n)nt

6. A formula that gives theamount accumulated byearning interest on principaland interest over time:

A (t) = P(1 + rn )nt .



Example 5

An investment of $500 is made in a 6-year CD that earns 4 12% annual interest

that is compounded monthly. How much will the CD be worth at the end of the6-year term?

Solution:

Here the principal P = $500, the interest rate r = 4 12 % = 0.045, and

because the interest is compounded monthly, n = 12. The investment ismodeled by the following,

To determine the amount in the account after 6 years evaluate A (6) andround off to the nearest cent.

Answer: The CD will be worth $654.65 at the end of the 6-year term.

Next we explore the effects of increasing n in the formula. For the sake of clarity welet P and r equal 1 and calculate accordingly.

A (t) = 500(1 +0.04512 )

12t

A (6)===

500(1 +0.04512 )

12(6)

500(1.00375)72654.65



Annual compounding (1 + 1n )n

Yearly (n = 1) (1 + 11 )1 = 2

Semi-annually (n = 2) (1 + 12 )2 = 2.25

Quarterly (n = 4) (1 + 14 )4 ≈ 2.44140

Monthly (n = 12) (1 + 112 )12 ≈ 2.61304

Weekly (n = 52) (1 + 152 )52 ≈ 2.69260

Daily (n = 365) (1 + 1365 )365 ≈ 2.71457

Hourly (n = 8760) (1 + 18760 )8760 ≈ 2.71813

Continuing this pattern, as n increases to say compounding every minute or evenevery second, we can see that the result tends toward the natural basee ≈ 2.71828.Compounding interest every instant leads to the continuouslycompounding interest formula7,7. A formula that gives the

amount accumulated byearning continuouslycompounded interest:A (t) = Pert.



Here P represents the initial principal amount invested, r represents the annualinterest rate, and t represents the time in years the investment is allowed to accruecontinuously compounded interest.

Example 6

An investment of $500 is made in a 6-year CD that earns 4 12% annual interest

that is compounded continuously. How much will the CD be worth at the end ofthe 6-year term?

Solution:

Here the principal P = $500, and the interest rate r = 4 12 % = 0.045.Since

the interest is compounded continuously we will use the formula A (t) = Pert.The investment is modeled by the following,

To determine the amount in the account after 6 years, evaluate A (6) andround off to the nearest cent.

Answer: The CD will be worth $654.98 at the end of the 6-year term.

A (t) = Pert

A (t) = 500e0.045t

A (6)===

500e0.045(6)500e0.27

654.98



Compare the previous two examples and note that compounding continuously maynot be as beneficial as it sounds. While it is better to compound interest more often,the difference is not that profound. Certainly, the interest rate is a much greaterfactor in the end result.

Try this! How much will a $1,200 CD, earning 5.2% annual interest compoundedcontinuously, be worth at the end of a 10-year term?

Answer: $2,018.43


KEY TAKEAWAYS

• Exponential functions have definitions of the form f (x) = bx where

b > 0 and b ≠ 1. The domain consists of all real numbers (−∞, ∞)and the range consists of positive numbers (0, ∞) . Also, all

exponential functions of this form have a y-intercept of (0, 1) and areasymptotic to the x-axis.

• If the base of an exponential function is greater than 1 (b > 1), then itsgraph increases or grows as it is read from left to right.

• If the base of an exponential function is a proper fraction (0 < b < 1),then its graph decreases or decays as it is read from left to right.

• The number 10 is called the common base and the number e is called thenatural base.

• The natural exponential function defined by f (x) = ex has a graphthat is very similar to the graph of g (x) = 3x .

• Exponential functions are one-to-one.



http://www.youtube.com/v/1alUOjTq5wc

TOPIC EXERCISES

PART A : EXPONENTIAL FUNCTIONS

Evaluate.

1. f (x) = 3x where f (−2), f (0), and f (2) .

2. f (x) = 10x where f (−1), f (0), and f (1) .

3. g (x) = ( 13 )x where g (−1) , g (0), and g (3) .

4. g (x) = ( 34 )x where g (−2) , g (−1) , and g (0) .

5. h (x) = 9−x where h (−1) , h (0), and h ( 12 ) .6. h (x) = 4−x where h (−1) , h (− 1

2 ), and h (0) .

7. f (x) = −2x + 1 where f (−1), f (0), and f (3) .

8. f (x) = 2 − 3x where f (−1), f (0), and f (2) .

9. g (x) = 10−x + 20 where g (−2) , g (−1) , and g (0) .

10. g (x) = 1 − 2−x where g (−1) , g (0), and g (1) .

Use a calculator to approximate the following to the nearesthundredth.

11. f (x) = 2x + 5where f (2.5) .12. f (x) = 3x − 10 where f (3.2) .

13. g (x) = 4x where g ( 2⎯⎯

√ ) .14. g (x) = 5x − 1 where g ( 3

⎯⎯√ ) .

15. h (x) = 10x where h (π) .

16. h (x) = 10x + 1 where h ( π3 )17. f (x) = 10−x − 2where f (1.5) .



18. f (x) = 5−x + 3where f (1.3) .

19. f (x) = ( 23 )x + 1where f (−2.7) .

20. f (x) = ( 35 )−x − 1where f (1.4) .

Sketch the function and determine the domain and range. Draw thehorizontal asymptote with a dashed line.

21. f (x) = 4x

22. g (x) = 3x

23. f (x) = 4x + 2

24. f (x) = 3x − 6

25. f (x) = 2x−2

26. f (x) = 4x+2

27. f (x) = 3x+1 − 4

28. f (x) = 10x−4 + 2

29. h (x) = 2x−3 − 2

30. h (x) = 3x+2 + 4

31.

32.

33.

34.

35. g (x) = 2−x − 3

36. g (x) = 3−x + 1

37. f (x) = 6 − 10−x

38. g (x) = 5 − 4−x

f (x) = ( 14 )

x

h (x) = ( 13 )

x

f (x) = ( 14 )

x

− 2

h (x) = ( 13 )

x

+ 2



39. f (x) = 5 − 2x

40. f (x) = 3 − 3x

PART B : NATURAL BASE E

Find f (−1), f (0), and f ( 32 ) for the given function. Use a calculator

where appropriate to approximate to the nearest hundredth.

41. f (x) = ex + 242. f (x) = ex − 443. f (x) = 5 − 3ex

44. f (x) = e−x + 345. f (x) = 1 + e−x

46. f (x) = 3 − 2e−x

47. f (x) = e−2x + 2

48. f (x) = e−x2 − 1

Sketch the function and determine the domain and range. Draw thehorizontal asymptote with a dashed line.

49. f (x) = ex − 350. f (x) = ex + 2

51. f (x) = ex+1

52. f (x) = ex−3

53. f (x) = ex−2 + 1

54. f (x) = ex+2 − 155. g (x) = −ex

56. g (x) = e−x

57. h (x) = −ex+1

58. h (x) = −ex + 3



PART C : COMPOUND INTEREST FORMULAS

59. Jim invested $750 in a 3-year CD that earns 4.2% annual interest that iscompounded monthly. How much will the CD be worth at the end of the 3-yearterm?

60. Jose invested $2,450 in a 4-year CD that earns 3.6% annual interest that iscompounded semi-annually. How much will the CD be worth at the end of the4-year term?

61. Jane has her $5,350 savings in an account earning 3 58 % annual interest that is

compounded quarterly. How much will be in the account at the end of 5 years?

62. Bill has $12,400 in a regular savings account earning 4 23 % annual interest that

is compounded monthly. How much will be in the account at the end of 3years?

63. If $85,200 is invested in an account earning 5.8% annual interest compoundedquarterly, then how much interest is accrued in the first 3 years?

64. If $124,000 is invested in an account earning 4.6% annual interest compoundedmonthly, then how much interest is accrued in the first 2 years?

65. Bill invested $1,400 in a 3-year CD that earns 4.2% annual interest that iscompounded continuously. How much will the CD be worth at the end of the3-year term?

66. Brooklyn invested $2,850 in a 5-year CD that earns 5.3% annual interest that iscompounded continuously. How much will the CD be worth at the end of the5-year term?

67. Omar has his $4,200 savings in an account earning 4 38 % annual interest that

is compounded continuously. How much will be in the account at the end of

2 12 years?

68. Nancy has her $8,325 savings in an account earning 5 78 % annual interest that

is compounded continuously. How much will be in the account at the end of

5 12 years?

69. If $12,500 is invested in an account earning 3.8% annual interest compoundedcontinuously, then how much interest is accrued in the first 10 years?

70. If $220,000 is invested in an account earning 4.5% annual interest compoundedcontinuously, then how much interest is accrued in the first 2 years?



71. The population of a certain small town is growing according to the function

P (t) = 12,500(1.02) t where t represents time in years since the lastcensus. Use the function to determine the population on the day of the census(when t = 0) and estimate the population in 6 years from that time.

72. The population of a certain small town is decreasing according to the function

P (t) = 22,300(0.982) t where t represents time in years since the lastcensus. Use the function to determine the population on the day of the census(when t = 0) and estimate the population in 6 years from that time.

73. The decreasing value, in dollars, of a new car is modeled by the formula

V (t) = 28,000(0.84) t where t represents the number of years after thecar was purchased. Use the formula to determine the value of the car when itwas new (t = 0) and the value after 4 years.

74. The number of unique visitors to the college website can be approximated by

the formula N (t) = 410(1.32) t where t represents the number of yearsafter 1997 when the website was created. Approximate the number of uniquevisitors to the college website in the year 2020.

75. If left unchecked, a new strain of flu virus can spread from a single person toothers very quickly. The number of people affected can be modeled using the

formula P (t) = e0.22t where t represents the number of days the virus isallowed to spread unchecked. Estimate the number of people infected with thevirus after 30 days and after 60 days.

76. If left unchecked, a population of 24 wild English rabbits can grow according to

the formula P (t) = 24e0.19t where the time t is measured in months. How

many rabbits would be present 3 12 years later?

77. The population of a certain city in 1975 was 65,000 people and was growingexponentially at an annual rate of 1.7%. At the time, the population growth

was modeled by the formula P (t) = 65,000e0.017t where t represented thenumber of years since 1975. In the year 2000, the census determined that theactual population was 104,250 people. What population did the model predictfor the year 2000 and what was the actual error?

78. Because of radioactive decay, the amount of a 10 milligram sample of

Iodine-131 decreases according to the formula A (t) = 10e−0.087t where trepresents time measured in days. How much of the sample remains after 10days?

79. The number of cells in a bacteria sample is approximated by the logistic

growth model N (t) = 1.2×105

1+9e−0.32t where t represents time in hours.



Determine the initial number of cells and then determine the number of cells 6hours later.

80. The market share of a product, as a percentage, is approximated by the

formula P (t) = 1002+e−0.44t where t represents the number of months after an

aggressive advertising campaign is launched. By how much can we expect themarket share to increase after the first three months of advertising?

PART D : D ISCUSSION BOARD

81. Why is b = 1 excluded as a base in the definition of exponential functions?Explain.

82. Explain why an exponential function of the form y = bx can never benegative.

83. Research and discuss the derivation of the compound interest formula.

84. Research and discuss the logistic growth model. Provide a link to moreinformation on this topic.

85. Research and discuss the life and contributions of Leonhard Euler.



ANSWERS

1. f (−2) = 19 , f (0) = 1, f (2) = 9

3. g (−1) = 3, g (0) = 1, g (3) = 127

5. h (−1) = 9, h (0) = 1, h ( 12 ) = 13

7. f (−1) = 12 , f (0) = 0, f (3) = −7

9. g (−2) = 120 , g (−1) = 30 , g (0) = 2111. 10.66

13. 7.10

15. 1385.46

17. −1.97

19. 3.99

21. Domain: (−∞, ∞) ; Range: (0, ∞)



23. Domain: (−∞, ∞) ; Range: (2, ∞)

25. Domain: (−∞, ∞) ; Range: (0, ∞)

27.



Domain: (−∞, ∞) ; Range: (−4, ∞)

29. Domain: (−∞, ∞) ; Range: (−2, ∞)

31. Domain: (−∞, ∞) ; Range: (0, ∞)



33. Domain: (−∞, ∞) ; Range: (−2, ∞)

35. Domain: (−∞, ∞) ; Range: (−3, ∞)



37. Domain: (−∞, ∞) ; Range: (−∞, 6)

39. Domain: (−∞, ∞) ; Range: (−∞, 5)41. f (−1) ≈ 2.37 , f (0) = 3, f ( 32 ) ≈ 6.48

43. f (−1) ≈ 3.90 , f (0) = 2, f ( 32 ) ≈ −8.45

45. f (−1) ≈ 3.72 , f (0) = 2, f ( 32 ) ≈ 1.22

47. f (−1) ≈ 9.39 , f (0) = 3, f ( 32 ) ≈ 2.05



49. Domain: (−∞, ∞) ; Range: (−3, ∞)

51. Domain: (−∞, ∞) ; Range: (0, ∞)

53.



Domain: (−∞, ∞) ; Range: (1, ∞)

55. Domain: (−∞, ∞) ; Range: (−∞, 0)

57. Domain: (−∞, ∞) ; Range: (−∞, 0)59. $850.52

61. $6,407.89

63. $16,066.13

65. $1,588.00

67. $4,685.44

69. $5,778.56



71. Initial population: 12,500; Population 6 years later: 14,077

73. New: $28,000; In 4 years: $13,940.40

75. After 30 days: 735 people; After 60 days: 540,365 people

77. Model: 99,423 people; error: 4,827 people

79. Initially there are 12,000 cells and 6 hours later there are 51,736 cells.

81. Answer may vary

83. Answer may vary

85. Answer may vary



7.3 Logarithmic Functions and Their Graphs

LEARNING OBJECTIVES

1. Define and evaluate logarithms.2. Identify the common and natural logarithm.3. Sketch the graph of logarithmic functions.

Definition of the Logarithm

We begin with the exponential function defined by f (x) = 2x and note that itpasses the horizontal line test.

Therefore it is one-to-one and has an inverse. Reflecting y = 2x about the liney = x we can sketch the graph of its inverse. Recall that if (x, y) is a point on the

graph of a function, then (y, x) will be a point on the graph of its inverse.


1639

To find the inverse algebraically, begin by interchanging x and y and then try tosolve for y.

We quickly realize that there is no method for solving for y. This function seems to“transcend” algebra. Therefore, we define the inverse to be the base-2 logarithm,denoted log2 x. The following are equivalent:

This gives us another transcendental function defined by f −1 (x) = log2 x, whichis the inverse of the exponential function defined by f (x) = 2x .

f (x)y

==2x

2x ⇒ x = 2y

y = log2 x ⇔ x = 2y


7.3 Logarithmic Functions and Their Graphs 1640

The domain consists of all positive real numbers (0, ∞) and the range consists of all

real numbers (−∞, ∞) . John Napier is widely credited for inventing the termlogarithm.

Figure 7.2

John Napier (1550–1617)

In general, given base b > 0 where b ≠ 1, the logarithm base b8 is defined asfollows:

Use this definition to convert logarithms to exponential form and back.

y = logb x if and only if x = by8. The exponent to which the

base b is raised in order toobtain a specific value. In otherwords, y = logb x isequivalent to by = x.



Logarithmic Form Exponential Form

log2 16 = 4 24 = 16

log5 25 = 2 52 = 25

log6 1 = 0 60 = 1

log3 3⎯⎯

√ = 12 31/2 = 3

⎯⎯√

log7 ( 149 ) = −2 7−2 = 1

49

It is useful to note that the logarithm is actually the exponent y to which the base bis raised to obtain the argument x.



Example 1

Evaluate:

a. log5 125b. log2 ( 18 )c. log4 2d. log11 1

Solution:

a. log5 125 = 3 because 53 = 125.b. log2 ( 18 ) = −3because 2−3 = 1

23 = 18 .

c. log4 2 = 12 because 41/2 = 4

⎯⎯√ = 2.

d. log11 1 = 0 because 110 = 1.

Note that the result of a logarithm can be negative or even zero. However, theargument of a logarithm is not defined for negative numbers or zero:

There is no power of two that results in −4 or 0. Negative numbers and zero are notin the domain of the logarithm. At this point it may be useful to go back and reviewall of the rules of exponents.

log2 (−4)log2 (0)

==??

⇒ ⇒

2?

2?==−40



Example 2

Find x:

a. log7 x = 2b. log16 x = 1

2c. log1/2 x = −5

Solution:

Convert each to exponential form and then simplify using the rules ofexponents.

a. log7 x = 2 is equivalent to 72 = x and thus x = 49.b. log16 x = 1

2 is equivalent to 161/2 = x or 16⎯ ⎯⎯⎯

√ = x and thus

x = 4.c. log1/2 x = −5 is equivalent to ( 12 )−5 = x or 25 = x and thus

x = 32

Try this! Evaluate: log5 ( 15√3 ).

Answer: − 13


The Common and Natural Logarithm

A logarithm can have any positive real number, other than 1, as its base. If the baseis 10, the logarithm is called the common logarithm9.

9. The logarithm base 10, denotedlog x.



http://www.youtube.com/v/1Bxv8PY91bw

When a logarithm is written without a base it is assumed to be the commonlogarithm. (Note: This convention varies with respect to the subject in which itappears. For example, computer scientists often let log x represent the logarithmbase 2.)

Example 3

Evaluate:

a. log 105

b. log 10⎯ ⎯⎯⎯

√c. log 0.01

Solution:

a. log 105 = 5 because 105 = 105 .b. log 10

⎯ ⎯⎯⎯√ = 1

2 because 101/2 = 10⎯ ⎯⎯⎯

√ .c. log 0.01 = log ( 1

100 ) = −2because

10−2 = 1102 = 1

100 = 0.01.

The result of a logarithm is not always apparent. For example, consider log 75.

f (x) = log10 x = log x Common logarithm

log 10log 75

log 100

===

1?2



We can see that the result of log 75 is somewhere between 1 and 2. On most

scientific calculators there is a common logarithm button LOG .Use it to find

the log 75 as follows:

Therefore, rounded off to the nearest thousandth, log 75 ≈ 1.875.As a check, wecan use a calculator to verify that 10 ^ 1.875 ≈ 75.

If the base of a logarithm is e, the logarithm is called the natural logarithm10.

The natural logarithm is widely used and is often abbreviated ln x.

LOG 75 = 1.87506

f (x) = loge x = ln x Natural logarithm

10. The logarithm base e, denotedln x.



Example 4

Evaluate:

a. ln e

b. ln e2⎯ ⎯⎯⎯

√3

c. ln ( 1e4 )

Solution:

a. ln e = 1 because ln e = loge e = 1and e1 = e.b. ln ( e2

⎯ ⎯⎯⎯√3 ) = 2

3because e2/3 = e2⎯ ⎯⎯⎯

√3 .

c. ln ( 1e4 ) = −4because e−4 = 1

e4

On a calculator you will find a button for the natural logarithm LN .

Therefore, rounded off to the nearest thousandth, ln (75) ≈ 4.317.As a check,we can use a calculator to verify that e ^ 4.317 ≈ 75.

LN 75 = 4.317488



Example 5

Find x. Round answers to the nearest thousandth.

a. log x = 3.2b. ln x = −4c. log x = − 2

3

Solution:

Convert each to exponential form and then use a calculator to approximate theanswer.

a. log x = 3.2 is equivalent to 103.2 = x and thus x ≈ 1584.893.b. ln x = −4 is equivalent to e−4 = x and thus x ≈ 0.018.c. log x = − 2

3 is equivalent to 10−2/3 = x and thus x ≈ 0.215

Try this! Evaluate: ln 1

e√

.

Answer: − 12


Graphing Logarithmic Functions

We can use the translations to graph logarithmic functions. When the base b > 1,the graph of f (x) = logb x has the following general shape:



http://www.youtube.com/v/fLW7_Zv7QME

The domain consists of positive real numbers, (0, ∞) and the range consists of all

real numbers, (−∞, ∞) . The y-axis, or x = 0, is a vertical asymptote and the x-

intercept is (1, 0) . In addition, f (b) = logb b = 1and so (b, 1) is a point on thegraph no matter what the base is.



Example 6

Sketch the graph and determine the domain and range:f (x) = log3 (x + 4) − 1.

Solution:

Begin by identifying the basic graph and the transformations.

Notice that the asymptote was shifted 4 units to the left as well. This defines thelower bound of the domain. The final graph is presented without theintermediate steps.

Answer:

y

y

y

===

log3 xlog3 (x + 4)log3 (x + 4) − 1

Basic graph

Shif t lef t 4 units.Shif t down 1 unit.



Domain: (−4, ∞); Range: (−∞, ∞)

Note: Finding the intercepts of the graph in the previous example is left for a latersection in this chapter. For now, we are more concerned with the general shape oflogarithmic functions.



Example 7

Sketch the graph and determine the domain and range:f (x) = − log (x − 2) .

Solution:


Here the vertical asymptote was shifted two units to the right. This defines thelower bound of the domain.

Answer:

y

y

y

===

log x

− log x

− log (x − 2)

Basic graph

Ref lection about the x-axisShif t right 2 units.



Domain: (2, ∞); Range: (−∞, ∞)

Try this! Sketch the graph and determine the domain and range:g (x) = ln (−x) + 2.

Answer:

Domain: (−∞, 0); Range: (−∞, ∞)


Next, consider exponential functions with fractional bases, such as the function

defined by f (x) = ( 12 )x . The domain consists of all real numbers. Choose some

values for x and then find the corresponding y-values.



http://www.youtube.com/v/6obhBJfCDe4

Use these points to sketch the graph and note that it passes the horizontal line test.

Therefore this function is one-to-one and has an inverse. Reflecting the graph aboutthe line y = x we have:

x

−2

−1

0

1

2

y

4

2

1

12

14

f (−2) = ( 12)

−2

= 22 = 4

f (−1) = ( 12)

−1

= 21 = 2

f (0) = ( 12 )

0

= 1

f (1) = ( 12 )

1

=12

f (2) = ( 12 )

2

=14

Solutions

(−2, 4)

(−1, 2)

(0, 1)

(1,12 )

(2,14 )



which gives us a picture of the graph of f −1 (x) = log1/2 x. In general, when thebase b > 1, the graph of the function defined by g (x) = log1/b x has the followingshape.

The domain consists of positive real numbers, (0, ∞) and the range consists of all

real numbers, (−∞, ∞) . The y-axis, or x = 0, is a vertical asymptote and the x-

intercept is (1, 0) . In addition, f (b) = log1/b b = −1and so (b, −1) is a point onthe graph.



Example 8

Sketch the graph and determine the domain and range:f (x) = log1/3 (x + 3) + 2.

Solution:


In this case the shift left 3 units moved the vertical asymptote to x = −3 whichdefines the lower bound of the domain.

Answer:

y

y

y

===

log1/3 xlog1/3 (x + 3)log1/3 (x + 3) + 2

Basic graph

Shif t lef t 3 units.Shif t up 2 units.



Domain: (−3, ∞); Range: (−∞, ∞)

In summary, if b > 1

And for both cases,

Domain :

Range :x-intercept :Asymptote :

(0, ∞)(−∞, ∞)(1, 0)x = 0



Try this! Sketch the graph and determine the domain and range:f (x) = log1/3 (x − 1) .

Answer:


KEY TAKEAWAYS

• The base-b logarithmic function is defined to be the inverse of the base-b

exponential function. In other words, y = logb x if and only ifby = x where b > 0 and b ≠ 1.

• The logarithm is actually the exponent to which the base is raised toobtain its argument.

• The logarithm base 10 is called the common logarithm and is denoted

log x.• The logarithm base e is called the natural logarithm and is denoted

ln x.• Logarithmic functions with definitions of the form f (x) = logb x

have a domain consisting of positive real numbers (0, ∞) and a range

consisting of all real numbers (−∞, ∞) . The y-axis, or x = 0 , is a

vertical asymptote and the x-intercept is (1, 0) .• To graph logarithmic functions we can plot points or identify the basic

function and use the transformations. Be sure to indicate that there is avertical asymptote by using a dashed line. This asymptote defines theboundary of the domain.



http://www.youtube.com/v/0oEVfInCAXw

TOPIC EXERCISES

PART A : DEFINITION OF THE LOGARITHM

Evaluate.

1. log3 9

2. log7 49

3. log4 4

4. log5 1

5. log5 625

6. log3 243

7.

8.

9.

10.

11. log4 410

12. log9 95

13. log5 5⎯⎯

√3

14. log2 2⎯⎯

√

15.

16.

log2 ( 116 )

log3 ( 19 )

log5 ( 1125 )

log2 ( 164 )

log7

1

7⎯⎯

√

log9 ( 19⎯⎯

√3 )



17. log1/2 4

18. log1/3 27

19.

20.

21. log25 5

22. log8 2

23.

24.

25.

26.

Find x.

27. log3 x = 4

28. log2 x = 5

29. log5 x = −3

30. log6 x = −2

31. log12 x = 0

32. log7 x = −1

33. log1/4 x = −2

34. log2/5 x = 2

35. log1/9 x = 12

36. log1/4 x = 32

37. log1/3 x = −1

log2/3 ( 23 )

log3/4 ( 916 )

log4 ( 12 )

log27 ( 13 )

log1/9 1

log3/5 ( 53 )



38. log1/5 x = 0

PART B : THE COMMON AND NATURAL LOGARITHM

Evaluate. Round off to the nearest hundredth where appropriate.

39. log 1000

40. log 100

41. log 0.1

42. log 0.0001

43. log 162

44. log 23

45. log 0.025

46. log 0.235

47. ln e4

48. ln 1

49.

50.

51. ln (25)52. ln (100)

53. ln (0.125)54. ln (0.001)

Find x. Round off to the nearest hundredth.

55. log x = 2.5

56. log x = 1.8

ln ( 1e )

ln ( 1e5 )



57. log x = −1.22

58. log x = −0.8

59. ln x = 3.1

60. ln x = 1.01

61. ln x = −0.69

62. ln x = −1

Find a without using a calculator.

63. log3 ( 127 ) = a

64. ln e = a

65. log2 a = 8

66. log2 2⎯⎯

√5 = a

67. log 1012 = a

68. ln a = 9

69. log1/8 ( 164 ) = a

70. log6 a = −3

71.

72.

In 1935 Charles Richter developed a scale used to measure earthquakes on aseismograph. The magnitude M of an earthquake is given by the formula,

ln a =15

log4/9 ( 23 ) = a

M = log ( I

I0 )



Here I represents the intensity of the earthquake as measured on theseismograph 100 km from the epicenter and I0 is the minimum intensityused for comparison. For example, if an earthquake intensity is measured tobe 100 times that of the minimum, then I = 100I0 and

The earthquake would be said to have a magnitude 2 on the Richter scale.Determine the magnitudes of the following intensities on the Richter scale.Round off to the nearest tenth.

73. I is 3 million times that of the minimum intensity.


75. I is the same as the minimum intensity.


In chemistry, pH is a measure of acidity and is given by the formula,

Here H+ represents the hydrogen ion concentration (measured in moles ofhydrogen per liter of solution.) Determine the pH given the followinghydrogen ion concentrations.

77. Pure water: H+ = 0.0000001

78. Blueberry: H+ = 0.0003162

79. Lemon Juice: H+ = 0.01

80. Battery Acid: H+ = 0.1

M = log ( 100I0I0 ) = log (100) = 2

pH = − log (H+)



PART C : GRAPHING LOGARITHMIC FUNCTIONS

Sketch the function and determine the domain and range. Draw thevertical asymptote with a dashed line.

81. f (x) = log2 (x + 1)

82. f (x) = log3 (x − 2)

83. f (x) = log2 x − 2

84. f (x) = log3 x + 3

85. f (x) = log2 (x − 2) + 4

86. f (x) = log3 (x + 1) − 2

87. f (x) = −log2 x + 1

88. f (x) = −log3 (x + 3)

89. f (x) = log2 (−x) + 1

90. f (x) = 2 − log3 (−x)

91. f (x) = log x + 5

92. f (x) = log x − 1

93. f (x) = log (x + 4) − 8

94. f (x) = log (x − 5) + 10

95. f (x) = − log (x + 2)

96. f (x) = − log (x − 1) + 2

97. f (x) = ln (x − 3)

98. f (x) = ln x + 3

99. f (x) = ln (x − 2) + 4

100. f (x) = ln (x + 5)101. f (x) = 2 − ln x



102. f (x) = − ln (x − 1)

103. f (x) = log1/2 x

104. f (x) = log1/3 x + 2

105. f (x) = log1/2 (x − 2)

106. f (x) = log1/3 (x + 1) − 1

107. f (x) = 2 − log1/4 x

108. f (x) = 1 + log1/4 (−x)

109. f (x) = 1 − log1/3 (x − 2)

110. f (x) = 1 + log1/2 (−x)

PART D : D ISCUSSION BOARD

111. Research and discuss the origins and history of the logarithm. How didstudents work with them before the common availability of calculators?

112. Research and discuss the history and use of the Richter scale. What does eachunit on the Richter scale represent?

113. Research and discuss the life and contributions of John Napier.



ANSWERS

1. 2

3. 1

5. 4

7. −4

9. −3

11. 10

13. 13

15. − 12

17. −2

19. 1

21. 12

23. − 12

25. 0

27. 81

29. 1125

31. 1

33. 16

35. 13

37. 3

39. 3

41. −1

43. 2.21

45. −1.60

47. 4



49. −1

51. 3.22

53. −2.08

55. 316.23

57. 0.06

59. 22.20

61. 0.50

63. −3

65. 256

67. 12

69. 2

71. e⎯⎯

√5

73. 6.5

75. 0

77. 7

79. 2

81. Domain: (−1, ∞) ; Range: (−∞, ∞)



83. Domain: (0, ∞) ; Range: (−∞, ∞)

85. Domain: (2, ∞) ; Range: (−∞, ∞)



87. Domain: (0, ∞) ; Range: (−∞, ∞)

89. Domain: (−∞, 0) ; Range: (−∞, ∞)



91. Domain: (0, ∞) ; Range: (−∞, ∞)

93. Domain: (−4, ∞) ; Range: (−∞, ∞)



95. Domain: (−2, ∞) ; Range: (−∞, ∞)

97. Domain: (3, ∞) ; Range: (−∞, ∞)



99. Domain: (2, ∞) ; Range: (−∞, ∞)

101. Domain: (0, ∞) ; Range: (−∞, ∞)



103. Domain: (0, ∞) ; Range: (−∞, ∞)

105. Domain: (2, ∞) ; Range: (−∞, ∞)



107. Domain: (0, ∞) ; Range: (−∞, ∞)

109. Domain: (2, ∞) ; Range: (−∞, ∞)111. Answer may vary




7.4 Properties of the Logarithm

LEARNING OBJECTIVES

1. Apply the inverse properties of the logarithm.2. Expand logarithms using the product, quotient, and power rule for

logarithms.3. Combine logarithms into a single logarithm with coefficient 1.

Logarithms and Their Inverse Properties

Recall the definition of the base-b logarithm: given b > 0 where b ≠ 1,

Use this definition to convert logarithms to exponential form. Doing this, we canderive a few properties:

y = logb x if and only if x = by

logb 1 = 0 because b0 = 1logb b = 1 because b1 = b

logb ( 1b) = −1 because b−1 =

1b


1675

Example 1

Evaluate:

a. log 1b. ln e

c. log5 ( 15 )Solution:

a. When the base is not written, it is assumed to be 10. This is thecommon logarithm,

b. The natural logarithm, by definition, has base e,

c. Because 5−1 = 15 we have,

Furthermore, consider fractional bases of the form 1/bwhere b > 1.

log 1 = log10 1 = 0

ln e = loge e = 1

log5 ( 15) = −1

log1/b b = −1 because ( 1b)

−1

=1−1

b−1=

b

1= b


7.4 Properties of the Logarithm 1676

Example 2

Evaluate:

a. log1/4 4b. log2/3 ( 32 )

Solution:

a. log1/4 4 = −1 because ( 14 )−1 = 4

b. log2/3 ( 32 ) = −1 because ( 23 )−1 = 32

Given an exponential function defined by f (x) = bx , where b > 0 and b ≠ 1, its

inverse is the base-b logarithm, f −1 (x) = logb x. And because f (f −1 (x)) = x

and f −1 (f (x)) = x, we have the following inverse properties of the logarithm11:

Since f −1 (x) = logb x has a domain consisting of positive values (0, ∞), the

property blogb x = x is restricted to values where x > 0.

f −1 (f (x))f (f −1 (x))

=

=

logb bx = x

blogb x = x ,

and

x > 0

11. Given b > 0 we havelogb b

x = x and

blogb x = x when x > 0.



Example 3

Evaluate:

a. log5 625b. 5log5 3

c. eln 5

Solution:

Apply the inverse properties of the logarithm.

a. log5 625 = log5 54 = 4

b. 5log5 3 = 3c. eln 5 = 5

In summary, when b > 0 and b ≠ 1, we have the following properties:

logb 1 = 0 logb b = 1

log1/b b = −1 logb ( 1b ) = −1

logb bx = x blogb x = x, x > 0



Try this! Evaluate: log 0.00001

Answer: −5


Product, Quotient, and Power Properties of Logarithms

In this section, three very important properties of the logarithm are developed.These properties will allow us to expand our ability to solve many more equations.We begin by assigning u and v to the following logarithms and then write them inexponential form:

Substitute x = bu and y = bv into the logarithm of a product logb (xy) and the

logarithm of a quotient logb ( xy ) .Then simplify using the rules of exponents and

the inverse properties of the logarithm.

Logarithm of a Product Logarithm of a Quotient

logb (xy)====

logb (bubv)logb b

u+v

u + v

logb x + logb y

logb ( xy )=

===

logb ( bu

bv )logb b

u−v

u − v

logb x − logb y

logb xlogb y

==u

v

⇒ ⇒

bu

bv==x

y



http://www.youtube.com/v/YFsZxKiMCTg

This gives us two essential properties: the product property of logarithms12,

and the quotient property of logarithms13,

In words, the logarithm of a product is equal to the sum of the logarithm of thefactors. Similarly, the logarithm of a quotient is equal to the difference of thelogarithm of the numerator and the logarithm of the denominator.

Example 4

Write as a sum: log2 (8x) .

Solution:

Apply the product property of logarithms and then simplify.

Answer: 3 + log2 x

logb (xy) = logb x + logb y

logb ( x

y ) = logb x − logb y

log2 (8x)===

log2 8 + log2 xlog2 2

3 + log2 x3 + log2 x

12. logb (xy) = logb x + logb y;the logarithm of a product isequal to the sum of thelogarithm of the factors.

13. logb ( xy ) = logb x − logb y;

the logarithm of a quotient isequal to the difference of thelogarithm of the numeratorand the logarithm of thedenominator.



Example 5

Write as a difference: log ( x10 ).

Solution:

Apply the quotient property of logarithms and then simplify.

Answer: log x − 1

Next we begin with logb x = u and rewrite it in exponential form. After raisingboth sides to the nth power, convert back to logarithmic form, and then backsubstitute.

This leads us to the power property of logarithms14,

log ( x

10)=

=

log x − log 10

log x − 1

logb x

logb xn

logb xn

=

==

u

nu

nlogb x

⇒

⇐

bu

(bu)nbnu

=

=

=

x

(x )n

xn

logb xn = nlogb x

14. logb xn = nlogb x; the

logarithm of a quantity raisedto a power is equal to thatpower times the logarithm ofthe quantity.



In words, the logarithm of a quantity raised to a power is equal to that power timesthe logarithm of the quantity.

Example 6

Write as a product:

a. log2 x4

b. log5 ( x⎯⎯

√ ) .Solution:

a. Apply the power property of logarithms.

b. Recall that a square root can be expressed using rationalexponents, x

⎯⎯√ = x1/2 . Make this replacement and then apply the

power property of logarithms.

In summary,

Product property of logarithms logb (xy) = logb x + logb y

log2 x4 = 4log2 x

log5 ( x⎯⎯

√ )==

log5 x1/2

12log5 x



Quotient property oflogarithms

logb ( xy ) = logb x − logb y

Power property of logarithms logb xn = nlogb x

We can use these properties to expand logarithms involving products, quotients,and powers using sums, differences and coefficients. A logarithmic expression iscompletely expanded when the properties of the logarithm can no further beapplied.

Caution: It is important to point out the following:

log (xy) ≠ log x ⋅ log y and log ( x

y ) ≠log x

log y



Example 7

Expand completely: ln (2x3) .Solution:

Recall that the natural logarithm is a logarithm base e, ln x = loge x.Therefore, all of the properties of the logarithm apply.

Answer: ln 2 + 3 ln x

ln (2x3)==

ln 2 + ln x3

ln 2 + 3 ln x

Product rule f or logarithms

Power rule f or logarithms



Example 8

Expand completely: log 10xy2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯√3 .

Solution:

Begin by rewriting the cube root using the rational exponent 13 and then apply

the properties of the logarithm.

Answer: 13 + 1

3 log x + 23 log y

log 10xy2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯√3 =

=

=

=

=

log (10xy2)1/313log (10xy2)

13 (log 10 + log x + log y2)13 (1 + log x + 2 log y)13

+13log x +

23log y



Example 9

Expand completely: log2 ( (x+1) 2

5y ).

Solution:

When applying the product property to the denominator, take care todistribute the negative obtained from applying the quotient property.

Answer: 2log2 (x + 1) − log2 5 − log2 y

Caution: There is no rule that allows us to expand the logarithm of a sum ordifference. In other words,

log2 ( (x + 1)2

5y )=

=

==

log2 (x + 1)2 − log2 (5y)log2 (x + 1)2 − (log2 5 + log2 y)log2 (x + 1)2 − log2 5 − log2 y2log2 (x + 1) − log2 5 − log2 y

Distribute.

log (x ± y) ≠ log x ± log y



Try this! Expand completely: ln 5y4

x√

.

Answer: ln 5 + 4 ln y − 12 ln x




http://www.youtube.com/v/h_Zu-SkATl0

Example 10

Given that log2 x = a, log2 y = b, and that log2 z = c, write the following interms of a, b and c:

a. log2 (8x2y)

b. log2

2x 4

z√

Solution:

a. Begin by expanding using sums and coefficients and then replace aand b with the appropriate logarithm.

b. Expand and then replace a, b, and c where appropriate.

log2 (8x2y)===

log2 8 + log2 x2 + log2 y

log2 8 + 2log2 x + log2 y3 + 2a + b

log2

2x4

z⎯⎯√

==

=

=

log2 (2x4) − log2 z1/2

log2 2 + log2 x4 − log2 z

1/2

log2 2 + 4log2 x −12log2 z

1 + 4a −12b



Next we will condense logarithmic expressions. As we will see, it is important to beable to combine an expression involving logarithms into a single logarithm withcoefficient 1. This will be one of the first steps when solving logarithmic equations.

Example 11

Write as a single logarithm with coefficient 1: 3log3 x − log3 y + 2log3 5.

Solution:

Begin by rewriting all of the logarithmic terms with coefficient 1. Use thepower rule to do this. Then use the product and quotient rules to simplifyfurther.

Answer: log3 ( 25x 3

y )

3log3x − log3y + 2log35 =

=

=

=

{log3x3 − log3y} + log35

2 quotient property

{log3 ( x 3

y ) + log325} product property

log3 ( x 3

y⋅ 25)

log3 ( 25x 3

y )



Example 12

Write as a single logarithm with coefficient 1: 12 ln x − 3 ln y − ln z.

Solution:

Begin by writing the coefficients of the logarithms as powers of their argument,after which we will apply the quotient rule twice working from left to right.

Answer: ln ( x√y3z )

12ln x − 3 ln y − ln z=

=

=

=

=

ln x1/2 − ln y3 − ln z

ln ( x1/2

y3 ) − ln z

ln ( x1/2

y3÷ z)

ln ( x1/2

y3⋅1z )

ln ( x1/2

y3z ) or =ln (x⎯⎯

√y3z )



Try this! Write as a single logarithm with coefficient1: 3 log (x + y) − 6 log z + 2 log 5.

Answer: log ( 25(x+y)3z6 )


KEY TAKEAWAYS

• Given any base b > 0 and b ≠ 1, we can say that logb 1 = 0,

logb b = 1, log1/b b = −1 and that logb ( 1b ) = −1.• The inverse properties of the logarithm are logb b

x = x and

blogb x = x where x > 0.• The product property of the logarithm allows us to write a product as a

sum: logb (xy) = logb x + logb y.• The quotient property of the logarithm allows us to write a quotient as a

difference: logb ( xy ) = logb x − logb y.

• The power property of the logarithm allows us to write exponents as

coefficients: logb xn = nlogb x.

• Since the natural logarithm is a base-e logarithm, ln x = loge x , all ofthe properties of the logarithm apply to it.

• We can use the properties of the logarithm to expand logarithmicexpressions using sums, differences, and coefficients. A logarithmicexpression is completely expanded when the properties of the logarithmcan no further be applied.

• We can use the properties of the logarithm to combine expressionsinvolving logarithms into a single logarithm with coefficient 1. This is anessential skill to be learned in this chapter.



http://www.youtube.com/v/o1AVbmS9AU8

TOPIC EXERCISES

PART A : LOGARITHMS AND THEIR INVERSE PROPERTIES

Evaluate:

1. log7 1

2. log1/2 2

3. log 1014

4. log 10−23

5. log3 310

6. log6 6

7. ln e7

8.

9.

10. log1/5 5

11. log3/4 ( 43 )12. log2/3 1

13. 2log2 100

14. 3log3 1

15. 10 log 18

16. eln 23

17. eln x 2

18. eln ex

ln ( 1e )

log1/2 ( 12 )



Find a:

19. ln a = 1

20. log a = −1

21. log9 a = −1

22. log12 a = 1

23. log2 a = 5

24. log a = 13

25. 2a = 726. ea = 23

27. loga 45 = 5

28. loga 10 = 1

PART B : PRODUCT, QUOTIENT, AND POWER PROPERTIESOF LOGARITHMS

Expand completely.

29. log4 (xy)30. log (6x)31. log3 (9x 2)32. log2 (32x 7)33. ln (3y2)34. log (100x 2)

35.

36.

log2 ( x

y2 )log5 ( 25

x )



37. log (10x 2y3)38. log2 (2x 4y5)

39.

40.

41.

42.

43. log6 [36(x + y) 4]44. ln [e4(x − y) 3]45. log7 (2 xy⎯ ⎯⎯⎯√ )46. ln (2x y⎯⎯√ )

47.

48.

49.

50.

51.

log3 ( x 3

yz2 )log ( x

y3z2 )log5 ( 1

x 2yz )log4 ( 1

16x 2z3 )

log3

x

2 y⎯⎯√3

z

log 2(x + y) 3

z2

log 100x 3

(y + 10) 3

log7

x

(y + z) 3⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√5

log5 ( x 3

yz2⎯ ⎯⎯⎯⎯⎯

√3 )



52.

Given log3 x = a, log3 y = b , and log3 z = c, write the followinglogarithms in terms of a, b, and c.

53.

54.

55.

56.

Given logb 2 = 0.43 , logb 3 = 0.68 , and logb 7 = 1.21 ,calculate the following. (Hint: Expand using sums, differences, andquotients of the factors 2, 3, and 7.)

57. logb 42

58. logb (36)59. logb ( 289 )60. logb 21

⎯ ⎯⎯⎯√

Expand using the properties of the logarithm and then approximateusing a calculator to the nearest tenth.

61. log (3.10 × 1025)62. log (1.40 × 10−33)63. ln (6.2e−15)64. ln (1.4e22)

Write as a single logarithm with coefficient 1.

65. log x + log y

log ( x 2

y3z2⎯ ⎯⎯⎯⎯⎯⎯

√5 )

log3 (27x 2y3z)log3 (xy3 z⎯⎯√ )log3 ( 9x 2y

z3 )log3 (

x⎯⎯

√3

yz2 )



66. log3 x − log3 y

67. log2 5 + 2log2 x + log2 y

68. log3 4 + 3log3 x + 12 log3 y

69. 3log2 x − 2log2 y + 12 log2 z

70. 4 log x − log y − log 2

71. log 5 + 3 log (x + y)72. 4log5 (x + 5) + log5 y

73. ln x − 6 ln y + ln z

74. log3 x − 2log3 y + 5log3 z

75. 7 log x − log y − 2 log z

76. 2 ln x − 3 ln y − ln z

77. 23 log3 x − 1

2 (log3 y + log3 z)78. 1

5 (log7 x + 2log7 y) − 2log7 (z + 1)

79. 1 + log2 x − 12 log2 y

80. 2 − 3log3 x + 13 log3 y

81. 13 log2 x + 2

3 log2 y

82. −2log5 x + 35 log5 y

83. − ln 2 + 2 ln (x + y) − ln z

84. −3 ln (x − y) − ln z + ln 5

85. 13 (ln x + 2 ln y) − (3 ln 2 + ln z)

86. 4 log 2 + 23 log x − 4 log (y + z)

87. log2 3 − 2log2 x + 12 log2 y − 4log2 z



88. 2log5 4 − log5 x − 3log5 y + 23 log5 z

Express as a single logarithm and simplify.

89. log (x + 1) + log (x − 1)

90. log2 (x + 2) + log2 (x + 1)

91. ln (x 2 + 2x + 1) − ln (x + 1)

92. ln (x 2 − 9) − ln (x + 3)

93. log5 (x 3 − 8) − log5 (x − 2)

94. log3 (x 3 + 1) − log3 (x + 1)

95. log x + log (x + 5) − log (x 2 − 25)96. log (2x + 1) + log (x − 3) − log (2x 2 − 5x − 3)



ANSWERS

1. 0

3. 14

5. 10

7. 7

9. 1

11. −1

13. 100

15. 18

17. x 2

19. e

21. 19

23. 25 = 32

25. log2 7

27. 4

29. log4 x + log4 y

31. 2 + 2log3 x

33. ln 3 + 2 ln y

35. log2 x − 2log2 y

37. 1 + 2 log x + 3 log y

39. 3log3 x − log3 y − 2log3 z

41. −2log5 x − log5 y − log5 z

43. 2 + 4log6 (x + y)45. log7 2 + 1

2 log7 x + 12 log7 y



47. 2log3 x + 13 log3 y − log3 z

49. 2 + 3 log x − 3 log (y + 10)51. 3log5 x − 1

3 log5 y − 23 log5 z

53. 3 + 2a + 3b + c

55. 2 + 2a + b − 3c57. 2.32

59. 0.71

61. log (3.1) + 25 ≈ 25.5

63. ln (6.2) − 15 ≈ −13.2

65. log (xy)67. log2 (5x 2y)

69.

71. log [5(x + y) 3]73.

75.

77.

79.

81.

log2 (x 3 z

⎯⎯√y2 )

ln ( xz

y6 )log ( x 7

yz2 )log3

x 2⎯ ⎯⎯⎯√3

yz⎯ ⎯⎯⎯√

log2

2x

y⎯⎯√

log2 ( xy2⎯ ⎯⎯⎯⎯⎯√3 )



83.

85.

87.

89. log (x 2 − 1)91. ln (x + 1)

93. log5 (x 2 + 2x + 4)95.

ln (x + y) 2

2z

ln (xy2⎯ ⎯⎯⎯⎯⎯√3

8z )log2 (

3 y⎯⎯√x 2z4 )

log ( x

x − 5 )



7.5 Solving Exponential and Logarithmic Equations

LEARNING OBJECTIVES

1. Solve exponential equations.2. Use the change of base formula to approximate logarithms.3. Solve logarithmic equations.

Solving Exponential Equations

An exponential equation15 is an equation that includes a variable as one of itsexponents. In this section we describe two methods for solving exponentialequations. First, recall that exponential functions defined by f (x) = bx whereb > 0 and b ≠ 1, are one-to-one; each value in the range corresponds to exactlyone element in the domain. Therefore, f (x) = f (y) implies x = y. The converse istrue because f is a function. This leads to the very important one-to-one propertyof exponential functions16:

Use this property to solve special exponential equations where each side can bewritten in terms of the same base.

bx = by if and only if x = y

15. An equation which includes avariable as an exponent.

16. Given b > 0 and b ≠ 1 wehave bx = by if and only ifx = y.


1701

Example 1

Solve: 32x−1 = 27.

Solution:

Begin by writing 27 as a power of 3.

Next apply the one-to-one property of exponential functions. In other words,set the exponents equal to each other and then simplify.

Answer: 2

32x−1

32x−1==2733

2x − 12xx

===

342


7.5 Solving Exponential and Logarithmic Equations 1702

Example 2

Solve: 161−3x = 2.

Solution:

Begin by writing 16 as a power of 2 and then apply the power rule forexponents.

Now that the bases are the same we can set the exponents equal to each otherand simplify.

Answer: 14

161−3x

(24)1−3x

24(1−3x)

=

=

=

2

2

21

4 (1 − 3x)4 − 12x

−12x

x

===

=

11−3−3−12

=14



Try this! Solve: 252x+3 = 125.

Answer: − 34


In many cases we will not be able to equate the bases. For this reason we develop asecond method for solving exponential equations. Consider the following equations:

We can see that the solution to 3x = 12 should be somewhere between 2 and 3. Agraphical interpretation follows.

To solve this we make use of fact that logarithms are one-to-one functions. Givenx, y > 0 the one-to-one property of logarithms17 follows:

32

3?

33

===

91227

logb x = logb y if and only if x = y

17. Given b > 0 and b ≠ 1where x, y > 0 we havelogb x = logb y if and onlyif x = y.



http://www.youtube.com/v/yU_tJg6Unnc

This property, as well as the properties of the logarithm, allows us to solveexponential equations. For example, to solve 3x = 12 apply the common logarithmto both sides and then use the properties of the logarithm to isolate the variable.

Approximating to four decimal places on a calculator.

An answer between 2 and 3 is what we expected. Certainly we can check by raising 3to this power to verify that we obtain a good approximation of 12.

Note that we are not multiplying both sides by “log”; we are applying the one-to-one property of logarithmic functions — which is often expressed as “taking the logof both sides.” The general steps for solving exponential equations are outlined inthe following example.

3x

log 3x

x log 3

x

===

=

12log 12log 12log 12log 3

One-to-one property of logarithmsPower rule f or logarithms

x = log (12) / log (3) ≈ 2.2619

3 ^ 2.2618 ≈ 12✓



Example 3

Solve: 52x−1 + 2 = 9.

Solution:

• Step 1: Isolate the exponential expression.

• Step 2: Take the logarithm of both sides. In this case, we will takethe common logarithm of both sides so that we can approximateour result on a calculator.

• Step 3: Apply the power rule for logarithms and then solve.

This is an irrational number which can be approximated using a calculator.Take care to group the numerator and the product in the denominator when

52x−1 + 252x−1

==97

log 52x−1 = log 7

log 52x−1

(2x − 1) log 52x log 5 − log 5

2x log 5

x

====

=

log 7log 7log 7log 5 + log 7log 5 + log 7

2 log 5

Distribute.



entering this into your calculator. To do this, make use of the parenthesis

buttons ( and ) :

Answer:log 5+log 7

2 log 5≈ 1.1045

x = (log 5 + log (7)) / (2 * log (5)) ≈ 1.1045



Example 4

Solve: e5x+3 = 1.

Solution:

The exponential function is already isolated and the base is e. Therefore, wechoose to apply the natural logarithm to both sides.

Apply the power rule for logarithms and then simplify.

Answer: − 35

On most calculators there are only two logarithm buttons, the common logarithm

LOG and the natural logarithm LN . If we want to approximate log3 10 we

have to somehow change this base to 10 or e. The idea begins by rewriting thelogarithmic function y = loga x , in exponential form.

e5x+3

ln e5x+3

==1ln 1

ln e5x+3

(5x + 3) ln e

(5x + 3) ⋅ 15x + 3

x

==

==

=

ln 1ln 1

00

−35

Recall ln e = 1 and ln1 = 0.



Here x > 0 and so we can apply the one-to-one property of logarithms. Apply thelogarithm base b to both sides of the function in exponential form.

And then solve for y.

Replace y into the original function and we have the very important change of baseformula18:

We can use this to approximate log3 10 as follows.

loga x = y ⇒ x = ay

x

logb x==ay

logb ay

logb xlogb xlogb a

=

=

ylogb a

y

loga x =logb xlogb a

log3 10 =log 10log 3

≈ 2.0959 or log3 10 =ln 10ln 3

≈ 2.0959

18. loga x = logb x

logb a; we can write

any base-a logarithm in termsof base-b logarithms using thisformula.



Notice that the result is independent of the choice of base. In words, we canapproximate the logarithm of any given base on a calculator by dividing thelogarithm of the argument by the logarithm of that given base.

Example 5

Approximate log7 120 the nearest hundredth.

Solution:

Apply the change of base formula and use a calculator.

On a calculator,

Answer: 2.46

log7 120 =log 120log 7

log (120) / log (7) ≈ 2.46



Try this! Solve: 23x+1 − 4 = 1. Give the exact and approximate answerrounded to four decimal places.

Answer:log 5−log 2

3 log 2≈ 0.4406


Solving Logarithmic Equations

A logarithmic equation19 is an equation that involves a logarithm with a variableargument. Some logarithmic equations can be solved using the one-to-one propertyof logarithms. This is true when a single logarithm with the same base can beobtained on both sides of the equal sign.

19. An equation that involves alogarithm with a variableargument.



http://www.youtube.com/v/kQf1-cEQUg8

Example 6

Solve: log2 (2x − 5) − log2 (x − 2) = 0.

Solution:

We can obtain two equal logarithms base 2 by adding log2 (x − 2) to both sidesof the equation.

Here the bases are the same and so we can apply the one-to-one property andset the arguments equal to each other.

Checking x = 3 in the original equation:

log2 (2x − 5) − log2 (x − 2)

log2 (2x − 5)=

=

0

log2 (x − 2)

log2 (2x − 5)2x − 5

x

===

log2 (x − 2)x − 23

log2 (2 (3) − 5)log2 1

0

===

log2 ((3) − 2)log2 10 ✓



Answer: 3

When solving logarithmic equations the check is very important becauseextraneous solutions can be obtained. The properties of the logarithm only applyfor values in the domain of the given logarithm. And when working with variablearguments, such as log (x − 2), the value of x is not known until the end of thisprocess. The logarithmic expression log (x − 2) is only defined for values x > 2.



Example 7

Solve: log (3x − 4) = log (x − 2) .

Solution:

Apply the one-to-one property of logarithms (set the arguments equal to eachother) and then solve for x.

When performing the check we encounter a logarithm of a negative number:

Try this on a calculator, what does it say? Here x = 1 is not in the domain oflog (x − 2) .Therefore our only possible solution is extraneous and weconclude that there are no solutions to this equation.

Answer: No solution, Ø.

Caution: Solving logarithmic equations sometimes leads to extraneous solutions —we must check our answers.

log (3x − 4)3x − 4

2xx

====

log (x − 2)x − 221

log (x − 2)==log (1 − 2)log (−1) Undef ined



Try this! Solve: ln (x2 − 15) − ln (2x) = 0.

Answer: 5


In many cases we will not be able to obtain two equal logarithms. To solve suchequations we make use of the definition of the logarithm. If b > 0, where b ≠ 1,then logb x = y implies that by = x. Consider the following common logarithmicequations (base 10),

We can see that the solution to log x = 0.5 will be somewhere between 1 and 10. Agraphical interpretation follows.

To find x we can apply the definition as follows.

log x

log x

log x

===

00.51

⇒ ⇒ ⇒

x

x

x

===

1?10

Because 100

Because 101

=

=

1.

10.

log10 x = 0.5 ⇒ 100.5 = x



http://www.youtube.com/v/-kBkk-q2pNs

This can be approximated using a calculator,

An answer between 1 and 10 is what we expected. Check this on a calculator.

x = 100.5 = 10 ^ 0.5 ≈ 3.1623

log 3.1623 ≈ 5✓



Example 8

Solve: log3 (2x − 5) = 2.

Solution:

Apply the definition of the logarithm.

Solve the resulting equation.

Check.

Answer: 7

log3 (2x − 5) = 2 ⇒ 2x − 5 = 32

2x − 52xx

===

9147

log3 (2 (7) − 5)log3 (9)

=?

=22 ✓



In order to apply the definition, we will need to rewrite logarithmic expressions as asingle logarithm with coefficient 1.The general steps for solving logarithmicequations are outlined in the following example.



Example 9

Solve: log2 (x − 2) + log2 (x − 3) = 1.

Solution:

• Step 1: Write all logarithmic expressions as a single logarithmwith coefficient 1. In this case, apply the product rule forlogarithms.

• Step 2: Use the definition and rewrite the logarithm inexponential form.

• Step 3: Solve the resulting equation. Here we can solve byfactoring.

• Step 4: Check. This step is required.

log2 (x − 2) + log2 (x − 3)log2 [(x − 2) (x − 3)]

==11

log2 [(x − 2) (x − 3)] = 1 ⇒(x − 2) (x − 3) = 21

(x − 2) (x − 3)x2 − 5x + 6x2 − 5x + 4

(x − 4) (x − 1)x − 4

x

======

220004

or x − 1x

==01



Check x = 4 Check x = 1

log2 (x − 2) + log2 (x − 3)

log2 (4 − 2) + log2 (4 − 3)log2 (2) + log2 (1)

1 + 0

=

===

1

111 ✓

log2 (x − 2) + log2 (x − 3)

log2 (1 − 2) + log2 (1 − 3)log2 (−1) + log2 (−2)

=

==

1

11 ✗

In this example, x = 1 is not in the domain of the given logarithmic expressionand is extraneous. The only solution is x = 4.

Answer: 4



Example 10

Solve: log (x + 15) − 1 = log (x + 6) .Solution:

Begin by writing all logarithmic expressions on one side and constants on theother.

Apply the quotient rule for logarithms as a means to obtain a single logarithmwith coefficient 1.

This is a common logarithm; therefore use 10 as the base when applying thedefinition.

log (x + 15) − 1

log (x + 15) − log (x + 6)=

=

log (x + 6)1

log (x + 15) − log (x + 6)log ( x + 15

x + 6 )=

=

1

1



Check.

Answer: −5

Try this! Solve: log2 (x) + log2 (x − 1) = 1.

Answer: 2


x + 15x + 6x + 15x + 15

−9xx

=

====

101

10 (x + 6)10x + 6045−5

log (x + 15) − 1

log (−5 + 15) − 1log 10 − 1

1 − 10

=

====

log (x + 6)log (−5 + 6)log 100 ✓



http://www.youtube.com/v/Lzwh1jkj-EY

Example 11

Find the inverse: f (x) = log2 (3x − 4) .

Solution:

Begin by replacing the function notation f (x) with y.

Interchange x and y and then solve for y.

The resulting function is the inverse of f. Present the answer using functionnotation.

Answer: f −1 (x) = 2x+43

f (x)y

==log2 (3x − 4)log2 (3x − 4)

x=log2 (3y − 4) ⇒ 3y − 43y

y

==

=

2x

2x + 42x + 4

3



KEY TAKEAWAYS

• If each side of an exponential equation can be expressed using the samebase, then equate the exponents and solve.

• To solve a general exponential equation, first isolate the exponentialexpression and then apply the appropriate logarithm to both sides. Thisallows us to use the properties of logarithms to solve for the variable.

• The change of base formula allows us to use a calculator to calculatelogarithms. The logarithm of a number is equal to the commonlogarithm of the number divided by the common logarithm of the givenbase.

• If a single logarithm with the same base can be isolated on each side ofan equation, then equate the arguments and solve.

• To solve a general logarithmic equation, first isolate the logarithm withcoefficient 1 and then apply the definition. Solve the resulting equation.

• The steps for solving logarithmic equations sometimes produceextraneous solutions. Therefore, the check is required.



TOPIC EXERCISES

PART A : SOLVING EXPONENTIAL EQUATIONS

Solve using the one-to-one property of exponential functions.

1. 3x = 81

2. 2−x = 16

3. 5x−1 = 25

4. 3x+4 = 27

5. 25x−2 = 16

6. 23x+7 = 8

7. 812x+1 = 3

8. 643x−2 = 2

9. 92−3x − 27 = 0

10. 81−5x − 32 = 0

11. 16x 2− 2 = 0

12. 4x 2−1 − 64 = 0

13. 9x(x+1) = 81

14. 4x(2x+5) = 64

15. 100 x 2 − 107x−3 = 0

16. e3(3x 2−1) − e = 0

Solve. Give the exact answer and the approximate answer rounded tothe nearest thousandth.

17. 3x = 5

18. 7x = 2



19. 4x = 9

20. 2x = 10

21. 5x−3 = 13

22. 3x+5 = 17

23. 72x+5 = 2

24. 35x−9 = 11

25. 54x+3 + 6 = 4

26. 107x−1 − 2 = 1

27. e2x−3 − 5 = 0

28. e5x+1 − 10 = 0

29. 63x+1 − 3 = 7

30. 8 − 109x+2 = 9

31. 15 − e3x = 2

32. 7 + e4x+1 = 1033. 7 − 9e−x = 434. 3 − 6e−x = 0

35. 5x2= 2

36. 32x 2−x = 1

37. 100e27x = 50

38. 6e12x = 2

39.

40.

Find the x- and y-intercepts of the given function.

31 + e−x

= 12

1 + 3e−x= 1



41. f (x) = 3x+1 − 4

42. f (x) = 23x−1 − 1

43. f (x) = 10x+1 + 2

44. f (x) = 104x − 5

45. f (x) = ex−2 + 1

46. f (x) = ex+4 − 4

Use a u-substitution to solve the following.

47. 32x − 3x − 6 = 0(Hint: Let u = 3x )

48. 22x + 2x − 20 = 0

49. 102x + 10x − 12 = 0

50. 102x − 10x − 30 = 0

51. e2x − 3ex + 2 = 0

52. e2x − 8ex + 15 = 0

Use the change of base formula to approximate the following to thenearest hundredth.

53. log2 5

54. log3 7

55.

56.

57. log1/2 10

58. log2/3 30

59. log2 5⎯⎯

√

60. log2 6⎯⎯

√3

log5 ( 23 )

log7 ( 15 )



61. If left unchecked, a new strain of flu virus can spread from a single person toothers very quickly. The number of people affected can be modeled using the

formula P (t) = e0.22t, where t represents the number of days the virus isallowed to spread unchecked. Estimate the number of days it will take 1,000people to become infected.

62. The population of a certain small town is growing according to the function

P (t) = 12,500(1.02) t , where t represents time in years since the lastcensus. Use the function to determine number of years it will take thepopulation to grow to 25,000 people.

PART B : SOLVING LOGARITHMIC EQUATIONS

Solve using the one-to-one property of logarithms.

63. log5 (2x + 4) = log5 (3x − 6)64. log4 (7x) = log4 (5x + 14)65. log2 (x − 2) − log2 (6x − 5) = 0

66. ln (2x − 1) = ln (3x)

67. log (x + 5) − log (2x + 7) = 0

68. ln (x 2 + 4x) = 2 ln (x + 1)

69. log3 2 + 2log3 x = log3 (7x − 3)

70. 2 log x − log 36 = 0

71. ln (x + 3) + ln (x + 1) = ln 8

72. log5 (x − 2) + log5 (x − 5) = log5 10

Solve.

73. log2 (3x − 7) = 5

74. log3 (2x + 1) = 2

75. log (2x + 20) = 1

76. log4 (3x + 5) = 12



77. log3 x2 = 2

78. log (x 2 + 3x + 10) = 1

79. ln (x 2 − 1) = 0

80. log5 (x 2 + 20) − 2 = 0

81. log2 (x − 5) + log2 (x − 9) = 5

82. log2 (x + 5) + log2 (x + 1) = 5

83. log4 x + log4 (x − 6) = 2

84. log6 x + log6 (2x − 1) = 2

85. log3 (2x + 5) − log3 (x − 1) = 2

86. log2 (x + 1) − log2 (x − 2) = 4

87. ln x − ln (x − 1) = 1

88. ln (2x + 1) − ln x = 2

89. 2log3 x = 2 + log3 (2x − 9)

90. 2log2 x = 3 + log2 (x − 2)

91. log2 (x − 2) = 2 − log2 x

92. log2 (x + 3) + log2 (x + 1) − 1 = 0

93. log x − log (x + 1) = 1

94. log2 (x + 2) + log2 (1 − x) = 1 + log2 (x + 1)

Find the x- and y-intercepts of the given function.

95. f (x) = log (x + 3) − 1

96. f (x) = log (x − 2) + 1

97. f (x) = log2 (3x) − 4

98. f (x) = log3 (x + 4) − 3



99. f (x) = ln (2x + 5) − 6

100. f (x) = ln (x + 1) + 2

Find the inverse of the following functions.

101. f (x) = log2 (x + 5)102. f (x) = 4 + log3 x

103. f (x) = log (x + 2) − 3

104. f (x) = ln (x − 4) + 1

105. f (x) = ln (9x − 2) + 5

106. f (x) = log6 (2x + 7) − 1

107. g (x) = e3x

108. g (x) = 10−2x

109. g (x) = 2x+3

110. g (x) = 32x + 5

111. g (x) = 10x+4 − 3

112. g (x) = e2x−1 + 1

Solve.

113. log (9x + 5) = 1 + log (x − 5)114. 2 + log2 (x 2 + 1) = log2 13

115. e5x−2 − e3x = 0

116. 3x 2 − 11 = 70

117. 23x − 5 = 0

118. log7 (x + 1) + log7 (x − 1) = 1

119. ln (4x − 1) − 1 = ln x



120. log (20x + 1) = log x + 2

121.

122. 2e−3x = 4

123. 2e3x = e4x+1

124. 2 log x + log x − 1 = 0

125. 3 log x = log (x − 2) + 2 log x

126. 2 ln 3 + ln x 2 = ln (x 2 + 1)127. In chemistry, pH is a measure of acidity and is given by the formula

pH = − log (H+), where H+ is the hydrogen ion concentration

(measured in moles of hydrogen per liter of solution.) Determine the hydrogenion concentration if the pH of a solution is 4.

128. The volume of sound, L in decibels (dB), is given by the formula

L = 10 log (I/10−12)where I represents the intensity of the sound in

watts per square meter. Determine the intensity of an alarm that emits 120 dBof sound.


129. Research and discuss the history and use of the slide rule.

130. Research and discuss real-world applications involving logarithms.

31 + e2x

= 2



ANSWERS

1. 4

3. 3

5. 65

7. − 38

9. 16

11. ± 12

13. −2, 1

15. 12 , 3

17.log 5log 3

≈ 1.465

19.log 3log 2

≈ 1.585

21.

23.

25. Ø

27.

29.

31.

33. ln 3 ≈ 1.099

35. ± log 2log 5

⎯ ⎯⎯⎯⎯⎯⎯⎯

√ ≈ ±0.656

37. − ln 227 ≈ −0.026

3 log 5 + log 13log 5

≈ 4.594

log 2 − 5 log 72 log 7

≈ −2.322

3 + ln 52

≈ 2.3051 − log 63 log 6

≈ 0.095

ln 133

≈ 0.855



39. − ln 2 ≈ −0.693

41. x-intercept: ( 2 log 2−log 3log 3

, 0); y-intercept: (0, −1)

43. x-intercept: None; y-intercept: (0, 12)

45. x-intercept: None; y-intercept: (0, 1+e2e2 )

47. 1

49. log 3

51. 0, ln 253. 2.32

55. −0.25

57. −3.32

59. 1.16

61. Approximately 31 days

63. 10

65. 35

67. −2

69. 12 , 3

71. 1

73. 13

75. −5

77. ±3

79. ± 2⎯⎯

√

81. 13

83. 8

85. 2



87.

89. 9

91. 1 + 5⎯⎯

√

93. Ø

95. x-intercept: (7, 0) ; y-intercept: (0, log 3 − 1)97. x-intercept: ( 163 , 0); y-intercept: None

99. x-intercept: ( e6−52 , 0); y-intercept: (0, ln 5 − 6)

101. f −1 (x) = 2x − 5

103. f −1 (x) = 10x+3 − 2

105.

107.

109. g−1 (x) = log2 x − 3

111. g−1 (x) = log (x + 3) − 4113. 55

115. 1

117.

119. 14−e

121.

123. ln 2 − 1125. Ø

127. 10−4 moles per liter


e

e − 1

f −1 (x) =ex−5 + 2

9g−1 (x) =

ln x

3

log2 53

ln (1/2)2



7.6 Applications

LEARNING OBJECTIVES

1. Use the compound and continuous interest formulas.2. Calculate doubling time.3. Use the exponential growth/decay model.4. Calculate the rate of decay given half-life.

Compound and Continuous Interest Formulas

Recall that compound interest occurs when interest accumulated for one period isadded to the principal investment before calculating interest for the next period.The amount A accrued in this manner over time t is modeled by the compoundinterest formula:

Here the initial principal P is accumulating compound interest at an annual rate rwhere the value n represents the number of times the interest is compounded in ayear.

A (t) = P(1 +r

n)nt


1735

Example 1

Susan invested $500 in an account earning 4 12% annual interest that is

compounded monthly.

a. How much will be in the account after 3 years?b. How long will it take for the amount to grow to $750?

Solution:

In this example, the principal P = $500, the interest rate r = 4 12 % = 0.045,

and because the interest is compounded monthly, n = 12. The investment canbe modeled by the following function:

a. Use this model to calculate the amount in the account after t = 3years.

Rounded off to the nearest cent, after 3 years, the amountaccumulated will be $572.12.

b. To calculate the time it takes to accumulate $750, set A (t) = 750and solve for t.

A (t)

A (t)

=

=

500(1 +0.04512 )

12t

500(1.00375)12t

A(3) ==≈

500(1.00375)12(3)

500(1.00375)36

572.12


7.6 Applications 1736

This results in an exponential equation that can be solved by firstisolating the exponential expression.

At this point take the common logarithm of both sides, apply thepower rule for logarithms, and then solve for t.

Using a calculator we can approximate the time it takes.

A(t) = 500(1.00375)12t

750 = 500(1.00375)12t

7507505001.5

=

=

=

500(1.00375)12t(1.00375)12t(1.00375)12t

log (1.5)log (1.5)

log (1.5)12 log (1.00375)

log (1.5)12 log (1.00375)

=

=

=

=

log (1.00375)12t12t log (1.00375)12 t log (1.00375)12 log (1.00375)

t

t = log (1.5) / (12 * log (1.00375)) ≈ 9 years



Answer:

a. $572.12b. Approximately 9 years.

The period of time it takes a quantity to double is called the doubling time20. Wenext outline a technique for calculating the time it takes to double an initialinvestment earning compound interest.

20. The period of time it takes aquantity to double.



Example 2

Mario invested $1,000 in an account earning 6.3% annual interest that iscompounded semi-annually. How long will it take the investment to double?

Solution:

Here the principal P = $1,000, the interest rate r = 6.3% = 0.063, andbecause the interest is compounded semi-annually n = 2. This investment canbe modeled as follows:

Since we are looking for the time it takes to double $1,000, substitute $2,000 forthe resulting amount A (t) and then solve for t.

At this point we take the common logarithm of both sides.

A (t)

A (t)

=

=

1,000(1 +0.0632 )

2t

1,000(1.0315)2t

2,0002,0001,000

2

=

=

=

1,000(1.0315)2t(1.0315)2t(1.0315)2t



Using a calculator we can approximate the time it takes:

Answer: Approximately 11.17 years to double at 6.3%.

If the investment in the previous example was one million dollars, how long wouldit take to double? To answer this we would use P = $1,000,000andA (t) = $2,000,000:

Dividing both sides by 1,000,000 we obtain the same exponential function as before.

2

log 2

log 2log 2

2 log (1.0315)

=

=

=

=

(1.0315)2tlog (1.0315)2t2t log (1.0315)t

t = log (2) / (2 * log (1.0315)) ≈ 11.17 years

A (t)

2,000,000

=

=

1,000(1.0315)2t1,000,000(1.0315)2t

2 = (1.0315)2t



Hence, the result will be the same, about 11.17 years. In fact, doubling time isindependent of the initial investment P.

Interest is typically compounded semi-annually (n = 2), quarterly (n = 4), monthly (n= 12), or daily (n = 365). However if interest is compounded every instant we obtaina formula for continuously compounding interest:

Here P represents the initial principal amount invested, r represents the annualinterest rate, and t represents the time in years the investment is allowed to accruecontinuously compounded interest.

A (t) = Pert



Example 3

Mary invested $200 in an account earning 5 34% annual interest that is

compounded continuously. How long will it take the investment to grow to$350?

Solution:

Here the principal P = $200and the interest rate

r = 5 34 % = 5.75% = 0.0575.Since the interest is compounded

continuously, use the formula A (t) = Pert.Hence, the investment can bemodeled by the following,

To calculate the time it takes to accumulate to $350, set A (t) = 350 and solvefor t.

Begin by isolating the exponential expression.

A (t) = 200e0.0575t

A (t)350

==200e0.0575t

200e0.0575t

350200

74

1.75

=

=

=

e0.0575t

e0.0575t

e0.0575t



Because this exponential has base e, we choose to take the natural logarithm ofboth sides and then solve for t.

Using a calculator we can approximate the time it takes:

Answer: It will take approximately 9.73 years.

When solving applications involving compound interest, look for the keyword“continuous,” or the keywords that indicate the number of annual compoundings.It is these keywords that determine which formula to choose.

Try this! Mario invested $1,000 in an account earning 6.3% annual interest thatis compounded continuously. How long will it take the investment to double?

Answer: Approximately 11 years.


ln (1.75)ln (1.75)ln (1.75)ln (1.75)0.0575

=

=

=

=

ln e0.0575t

0.0575t ln e

0.0575t ⋅ 1

t

Apply the power rule f or logarithms.

Recall that ln e = 1.

t = ln (1.75) /0.0575 ≈ 9.73 years



http://www.youtube.com/v/Z_ilJHIQvWQ

Modeling Exponential Growth and Decay

In the sciences, when a quantity is said to grow or decay exponentially, it isspecifically meant to be modeled using the exponential growth/decay formula21:

Here P0 , read “P naught,” or “P zero,” represents the initial amount, k representsthe growth rate, and t represents the time the initial amount grows or decaysexponentially. If k is negative, then the function models exponential decay. Noticethat the function looks very similar to that of continuously compounding interestformula. We can use this formula to model population growth when conditions areoptimal.

P (t) = P0ekt

21. A formula that modelsexponential growth or decay:

P (t) = P0ekt .



Example 4

It is estimated that the population of a certain small town is 93,000 people withan annual growth rate of 2.6%. If the population continues to increaseexponentially at this rate:

a. Estimate the population in 7 years’ time.b. Estimate the time it will take for the population to reach 120,000

people.

Solution:

We begin by constructing a mathematical model based on the giveninformation. Here the initial population P0 = 93,000 people and the growthrate r = 2.6% = 0.026.The following model gives population in terms of timemeasured in years:

a. Use this function to estimate the population in t = 7 years.

b. Use the model to determine the time it takes to reachP (t) = 120,000people.

P (t) = 93,000e0.026t

P(t)==≈

93,000e0.026(7)

93,000e0.182

111,564 people



Take the natural logarithm of both sides and then solve for t.

Using a calculator,

Answer:

a. 111,564 peopleb. 9.8 years

Often the growth rate k is not given. In this case, we look for some otherinformation so that we can determine it and then construct a mathematical model.The general steps are outlined in the following example.

P (t)120,000120,00093,000

4031

==

=

=

93,000e0.026t

93,000e0.026t

e0.026t

e0.026t

ln ( 4031)

ln ( 4031)

ln ( 4031)

ln ( 4031 )0.026

=

=

=

=

ln e0.026t

0.026t ln e

0.026t ⋅ 1

t

t = ln (40/31) /0.026 ≈ 9.8 years



Example 5

Under optimal conditions Escherichia coli (E. coli) bacteria will growexponentially with a doubling time of 20 minutes. If 1,000 E. coli cells are placedin a Petri dish and maintained under optimal conditions, how many E. coli cellswill be present in 2 hours?

Figure 7.3

Escherichia coli (E. coli) (Wikipedia)

Solution:

The goal is to use the given information to construct a mathematical modelbased on the formula P (t) = P0ekt .

• Step 1: Find the growth rate k. Use the fact that the initial amount,P0 = 1,000 cells, doubles in 20 minutes. That is, P (t) = 2,000cells when t = 20minutes.

Solve for the only variable k.

P (t)2,000

==P0e

kt

1,000ek20



• Step 2: Write a mathematical model based on the giveninformation. Here k ≈ 0.0347, which is about 3.5% growth rateper minute. However, we will use the exact value for k in ourmodel. This will allow us to avoid round-off error in the finalresult. Use P0 = 1,000 and k = ln (2) /20:

This equation models the number of E. coli cells in terms of time inminutes.

• Step 3: Use the function to answer the questions. In this case, weare asked to find the number of cells present in 2 hours. Becausetime is measured in minutes, use t = 120minutes to calculate thenumber of E. coli cells.

2,0002,0001,000

2ln (2)ln (2)ln (2)ln (2)20

=

=

====

=

1,000ek20

ek20

ek20

ln ek20

k20 ln e

k20 ⋅ 1

k

P (t) = 1,000e(ln (2)/20)t

P (120)=

=

===

1,000e(ln (2)/20)(120)1,000eln (2)⋅6

1,000eln 26

1,000 ⋅ 26

64,000 cells



Answer: In two hours 64,000 cells will be present.

When the growth rate is negative the function models exponential decay. We candescribe decreasing quantities using a half-life22, or the time it takes to decay toone-half of a given quantity.

22. The period of time it takes aquantity to decay to one-half ofthe initial amount.



Example 6

Due to radioactive decay, caesium-137 has a half-life of 30 years. How long willit take a 50-milligram sample to decay to 10 milligrams?

Solution:

Use the half-life information to determine the rate of decay k. In t = 30 yearsthe initial amount P0 = 50 milligrams will decay to half P (30) = 25milligrams.

Solve for the only variable, k.

P (t)25

==P0e

kt

50ek30

252550

ln ( 12 )

ln ( 12 )

ln 1 − ln 230

−ln 230

=

=

=

=

=

=

50ek30

e30k

ln e30k

30k ln e

k

k

Recall that ln 1 = 0.



Note that k = − ln 230 ≈ −0.0231is negative. However, we will use the exact

value to construct a model that gives the amount of cesium-137 with respect totime in years.

Use this model to find t when P (t) = 10milligrams.

Answer: Using a calculator, it will take t ≈ 69.66years to decay to 10milligrams.

Radiocarbon dating is a method used to estimate the age of artifacts based on therelative amount of carbon-14 present in it. When an organism dies, it stopsabsorbing this naturally occurring radioactive isotope, and the carbon-14 begins to

P (t) = 50e(− ln 2/30)t

101050

ln ( 15 )

ln 1 − ln 5

−30 (ln 1 − ln 5)

ln 2−30 (0 − ln 5)

ln 230 ln 5ln 2

=

=

=

=

=

=

=

50e(− ln 2/30)t

e(− ln 2/30)t

ln e(− ln 2/30)t

(−ln 230 ) t ln e

t

t

t

Recall that ln e = 1.



decay at a known rate. Therefore, the amount of carbon-14 present in an artifactcan be used to estimate the age of the artifact.



Example 7

An ancient bone tool is found to contain 25% of the carbon-14 normally foundin bone. Given that carbon-14 has a half-life of 5,730 years, estimate the age ofthe tool.

Solution:

Begin by using the half-life information to find k. Here the initial amount P0 ofcarbon-14 is not given, however, we know that in t = 5,730years, this amount

decays to half, 12 P0 .

Dividing both sides by P0 leaves us with an exponential equation in terms of k.This shows that half-life is independent of the initial amount.

Solve for k.

P (t)12P0

=

=

P0ekt

P0ek5,730

12

= ek5,730



Therefore we have the model,

Next we wish to the find time it takes the carbon-14 to decay to 25% of theinitial amount, or P (t) = 0.25P0 .

Divide both sides by P0 and solve for t.

ln ( 12 )

ln 1 − ln 20 − ln 25,730

−ln 25,730

=

=

=

=

ln ek5,730

5,730k ln e

k

k

P (t) = P0e(− ln 2/5,730)t

0.25P0 = P0e(− ln 2/5,730)t



Answer: The tool is approximately 11,460 years old.

Try this! The half-life of strontium-90 is about 28 years. How long will it take a36 milligram sample of strontium-90 to decay to 30 milligrams?

Answer: 7.4 years


0.25

ln (0.25)ln (0.25)

−5,730 ln (0.25)

ln 211,460

=

=

=

=

≈

e(− ln 2/5,730)tln e(− ln 2/5,730)t

(−ln 25,730) t ln e

t

t



http://www.youtube.com/v/Oto0ihiyvBc

KEY TAKEAWAYS

• When interest is compounded a given number of times per year use the

formula A (t) = P(1 + rn )nt .

• When interest is to be compounded continuously use the formulaA (t) = Pert.

• Doubling time is the period of time it takes a given amount to double.Doubling time is independent of the principal.

• When amounts are said to be increasing or decaying exponentially, use

the formula P (t) = P0ekt .• Half-life is the period of time it takes a given amount to decrease to one-

half. Half-life is independent of the initial amount.• To model data using the exponential growth/decay formula, use the

given information to determine the growth/decay rate k. Once k isdetermined, a formula can be written to model the problem. Use theformula to answer the questions.



TOPIC EXERCISES

PART A : COMPOUND AND CONTINUOUS INTEREST

1. Jill invested $1,450 in an account earning 4 58 % annual interest that is

compounded monthly.

a. How much will be in the account after 6 years?b. How long will it take the account to grow to $2,200?

2. James invested $825 in an account earning 5 25 % annual interest that is

compounded monthly.


3. Raul invested $8,500 in an online money market fund earning 4.8% annualinterest that is compounded continuously.


4. Ian deposited $500 in an account earning 3.9% annual interest that iscompounded continuously.


5. Bill wants to grow his $75,000 inheritance to $100,000 before spending any ofit. How long will this take if the bank is offering 5.2% annual interestcompounded quarterly?

6. Mary needs $25,000 for a down payment on a new home. If she invests hersavings of $21,350 in an account earning 4.6% annual interest that iscompounded semi-annually, how long will it take to grow to the amount thatshe needs?

7. Joe invested his $8,700 savings in an account earning 6 34 % annual interest

that is compounded continuously. How long will it take to earn $300 ininterest?

8. Miriam invested $12,800 in an account earning 5 14 % annual interest that is

compounded monthly. How long will it take to earn $1,200 in interest?



9. Given that the bank is offering 4.2% annual interest compounded monthly,what principal is needed to earn $25,000 in interest for one year?

10. Given that the bank is offering 3.5% annual interest compounded continuously,what principal is needed to earn $12,000 in interest for one year?

11. Jose invested his $3,500 bonus in an account earning 5 12 % annual interest

that is compounded quarterly. How long will it take to double his investment?

12. Maria invested her $4,200 savings in an account earning 6 34 % annual interest

that is compounded semi-annually. How long will it take to double her savings?

13. If money is invested in an account earning 3.85% annual interest that iscompounded continuously, how long will it take the amount to double?

14. If money is invested in an account earning 6.82% annual interest that iscompounded continuously, how long will it take the amount to double?

15. Find the annual interest rate at which an account earning continuouslycompounding interest has a doubling time of 9 years.

16. Find the annual interest rate at which an account earning interest that iscompounded monthly has a doubling time of 10 years.

17. Alice invested her savings of $7,000 in an account earning 4.5% annual interestthat is compounded monthly. How long will it take the account to triple invalue?

18. Mary invested her $42,000 bonus in an account earning 7.2% annual interestthat is compounded continuously. How long will it take the account to triple invalue?

19. Calculate the doubling time of an investment made at 7% annual interest thatis compounded:

a. monthlyb. continuously

20. Calculate the doubling time of an investment that is earning continuouslycompounding interest at an annual interest rate of:

a. 4%b. 6%

21. Billy’s grandfather invested in a savings bond that earned 5.5% annual interestthat was compounded annually. Currently, 30 years later, the savings bond isvalued at $10,000. Determine what the initial investment was.



22. In 1935 Frank opened an account earning 3.8% annual interest that wascompounded quarterly. He rediscovered this account while cleaning out hisgarage in 2005. If the account is now worth $11,294.30, how much was hisinitial deposit in 1935?

PART B : MODELING EXPONENTIAL GROWTH AND DECAY

23. The population of a small town of 24,000 people is expected grow exponentiallyat a rate of 1.6% per year. Construct an exponential growth model and use it to:

a. Estimate the population in 3 years’ time.b. Estimate the time it will take for the population to reach 30,000 people.

24. During the exponential growth phase, certain bacteria can grow at a rate of4.1% per hour. If 10,000 cells are initially present in a sample, construct anexponential growth model and use it to:

a. Estimate the population in 5 hours.b. Estimate the time it will take for the population to reach 25,000 cells.

25. In 2000, the world population was estimated to be 6.115 billion people and in2010 the estimate was 6.909 billion people. If the world population continues togrow exponentially, estimate the total world population in 2020.

26. In 2000, the population of the United States was estimated to be 282 millionpeople and in 2010 the estimate was 309 million people. If the population of theUnited States grows exponentially, estimate the population in 2020.

27. An automobile was purchased new for $42,500 and 2 years later it was valuedat $33,400. Estimate the value of the automobile in 5 years if it continues todecrease exponentially.

28. A new PC was purchased for $1,200 and in 1.5 years it was worth $520. Assumethe value is decreasing exponentially and estimate the value of the PC fouryears after it is purchased.

29. The population of the downtown area of a certain city decreased from 12,500people to 10,200 people in two years. If the population continues to decreaseexponentially at this rate, what would we expect the population to be in twomore years?

30. A new MP3 player was purchased for $320 and in 1 year it was selling usedonline for $210. If the value continues to decrease exponentially at this rate,determine the value of the MP3 player 3 years after it was purchased.



31. The half-life of radium-226 is about 1,600 years. How long will a 5-milligramsample of radium-226 take to decay to 1 milligram?

32. The half-life of plutonium-239 is about 24,000 years. How long will a5-milligram sample of plutonium-239 take to decay to 1 milligram?

33. The half-life of radioactive iodine-131 is about 8 days. How long will it take a28-gram initial sample of iodine-131 to decay to 12 grams?

34. The half-life of caesium-137 is about 30 years. How long will it take a15-milligram sample of caesium-137 to decay to 5 milligrams?

35. The Rhind Mathematical Papyrus is considered to be the best example ofEgyptian mathematics found to date. This ancient papyrus was found tocontain 64% of the carbon-14 normally found in papyrus. Given that carbon-14has a half-life of 5,730 years, estimate the age of the papyrus.

36. A wooden bowl artifact carved from oak was found to contain 55% of thecarbon-14 normally found in oak. Given that carbon-14 has a half-life of 5,730years, estimate the age of the bowl.

37. The half-life of radioactive iodine-131 is about 8 days. How long will it take asample of iodine-131 to decay to 10% of the original amount?

38. The half-life of caesium-137 is about 30 years. How long will it take a sample ofcaesium-137 to decay to 25% of the original amount?

39. The half-life of caesium-137 is about 30 years. What percent of an initialsample will remain in 100 years?

40. The half-life of radioactive iodine-131 is about 8 days. What percent of aninitial sample will remain in 30 days?

41. If a bone is 100 years old, what percent of its original amount of carbon-14 dowe expect to find in it?

42. The half-life of plutonium-239 is about 24,000 years. What percent of an initialsample will remain in 1,000 years?

43. Find the amount of time it will take for 10% of an initial sample ofplutonium-239 to decay. (Hint: If 10% decays, then 90% will remain.)

44. Find the amount of time it will take for 10% of an initial sample of carbon-14 todecay.

Solve for the given variable:

45. Solve for t: A = Pert



46. Solve for t: A = P(1 + r)t

47. Solve for I: M = log ( II0 )

48. Solve for H+ : pH = − log (H+)49. Solve for t: P = 1

1+e−t

50. Solve for I: L = 10 log (I/10−12)51. The number of cells in a certain bacteria sample is approximated by the

logistic growth model N (t) = 1.2×105

1+9e−0.32t , where t represents time in hours.

Determine the time it takes the sample to grow to 24,000 cells.

52. The market share of a product, as a percentage, is approximated by the

formula P (t) = 1003+e−0.44t where t represents the number of months after an

aggressive advertising campaign is launched.

a. What was the initial market share?b. How long would we expect to see a 3.5% increase in market share?

53. In chemistry, pH is a measure of acidity and is given by the formula

pH = − log (H+), where H+ is the hydrogen ion concentration

(measured in moles of hydrogen per liter of solution.) What is the hydrogenion concentration of seawater with a pH of 8?

54. Determine the hydrogen ion concentration of milk with a pH of 6.6.

55. The volume of sound, L in decibels (dB), is given by the formula

L = 10 log (I/10−12)where I represents the intensity of the sound in

watts per square meter. Determine the sound intensity of a hair dryer thatemits 70 dB of sound.

56. The volume of a chainsaw measures 110 dB. Determine the intensity of thissound.


57. Which factor affects the doubling time the most, the annual compounding n orthe interest rate r? Explain.



58. Research and discuss radiocarbon dating. Post something interesting you havelearned as well as a link to more information.

59. Is exponential growth sustainable over an indefinite amount of time? Explain.

60. Research and discuss the half-life of radioactive materials.



ANSWERS

1. a. $1,912.73b. 9 years

3. a. $9,356.45b. 3.4 years

5. 5.6 years

7. 12 year

9. $583,867

11. 12.7 years

13. 18 years

15. 7.7%

17. 24.5 years

19. a. 9.93 yearsb. 9.90 years

21. $2,006.44

23. a. About 25,180 peopleb. About 14 years

25. About 7.806 billion people

27. About $23,269.27

29. 8,323 people

31. 3,715 years

33. 9.8 days

35. About 3,689 years old

37. 26.6 days

39. 9.9%

41. 98.8%

43. 3,648 years



45.

47. I = I0 ⋅ 10M

49.

51. Approximately 2.5 hours

53. 10−8 moles per liter

55. 10−5 watts per square meter

57. Answer may vary

59. Answer may vary

t =ln (A) − ln (P)

r

t = ln ( P

1 − P)



7.7 Review Exercises and Sample Exam


1765

REVIEW EXERCISES

COMPOSITION AND INVERSE FUNCTIONS

Given f and g find (f○g) (x) and (g○f ) (x) .1. f (x) = 6x − 5 , g (x) = 2x + 1

2. f (x) = 5 − 6x , g (x) = 32 x

3. f (x) = 2x 2 + x − 2 , g (x) = 5x

4. f (x) = x 2 − x − 6, g (x) = x − 3

5. f (x) = x + 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√ , g (x) = 8x − 2

6. f (x) = x−13x−1 , g (x) = 1

x

7. f (x) = x 2 + 3x − 1 , g (x) = 1x−2

8. f (x) = 3 (x + 2)⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3 , g (x) = 9x 3 − 2

Are the given functions one-to-one? Explain.

9.


7.7 Review Exercises and Sample Exam 1766

10.

11.

12.



Verify algebraically that the two given functions are inverses. In other

words, show that (f○f −1) (x) = x and (f −1○f) (x) = x.

13. f (x) = 6x − 5 , f −1 (x) = 16 x + 5

6

14. f (x) = 2x + 3⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√ , f −1 (x) = x 2−32 , x ≥ 0

15. f (x) = x3x−2 , f −1 (x) = 2x

3x−1

16. f (x) = x + 3⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√3 − 4, f −1 (x) = (x + 4)3 − 3

Find the inverses of each function defined as follows:

17. f (x) = −7x + 3

18. f (x) = 23 x − 1

2

19. g (x) = x 2 − 12, x ≥ 0

20. g (x) = (x − 1)3 + 5

21. g (x) = 2x−1

22. h (x) = x+5x−5

23. h (x) = 3x−1x

24. p (x) = 5x⎯ ⎯⎯⎯

√3 + 3

25. h (x) = 2x − 7⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√3 + 2

26. h (x) = x + 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√5 − 3

EXPONENTIAL FUNCTIONS AND THEIR GRAPHS

Evaluate.

27. f (x) = 5x ; find f (−1), f (0), and f (3) .

28. f (x) = ( 12 )x ; find f (−4), f (0), and f (−3) .

29. g (x) = 10−x ; find g (−5) , g (0), and g (2) .



30. g (x) = 1 − 3x ; find g (−2) , g (0), and g (3) .

Sketch the exponential function. Draw the horizontal asymptote with adashed line.

31. f (x) = 5x + 10

32. f (x) = 5x−4

33. f (x) = −3x − 9

34. f (x) = 3x+2 + 6

35.

36.

37. f (x) = 2−x + 3

38. f (x) = 1 − 3−x

Use a calculator to evaluate the following. Round off to the nearesthundredth.

39. f (x) = ex + 1; find f (−3), f (−1), and f ( 12 ) .40. g (x) = 2 − 3ex ; find g (−1) , g (0), and g ( 23 ) .41. p (x) = 1 − 5e−x ; find p (−4) , p (− 1

2 ), and p (0) .

42. r (x) = e−2x − 1; find r (−1), r ( 14 ), and r (2) .

Sketch the function. Draw the horizontal asymptote with a dashed line.

43. f (x) = ex + 4

44. f (x) = ex−4

45. f (x) = ex+3 + 246. f (x) = e−x + 5

47. Jerry invested $6,250 in an account earning 3 58 % annual interest that is

compounded monthly. How much will be in the account after 4 years?

f (x) = ( 13 )

x

f (x) = ( 12 )

x

− 4



48. Jose invested $7,500 in an account earning 4 14 % annual interest that is

compounded continuously. How much will be in the account after 3 12 years?

49. A 14-gram sample of radioactive iodine is accidently released into theatmosphere. The amount of the substance in grams is given by the formula

P (t) = 14e−0.087t , where t represents the time in days after the sample wasreleased. How much radioactive iodine will be present in the atmosphere 30days after it was released?

50. The number of cells in a bacteria sample is given by the formula

N (t) = 2.4×105

1+9e−0.28t , where t represents the time in hours since the initial

placement of 24,000 cells. Use the formula to calculate the number of cells inthe sample 20 hours later.

LOGARITHMIC FUNCTIONS AND THEIR GRAPHS

Evaluate.

51. log4 16

52. log3 27

53.

54.

55. log1/3 9

56. log3/4 ( 43 )57. log7 1

58. log3 (−3)

59. log4 0

60. log3 81

61. log6 6⎯⎯

√

62. log5 25⎯ ⎯⎯⎯

√3

log2 ( 132 )

log ( 110 )



63. ln e8

64.

65. log (0.00001)

66. log 1,000,000

Find x.

67. log5 x = 3

68. log3 x = −4

69. log2/3 x = 3

70. log3 x = 25

71. log x = −3

72.

Sketch the graph of the logarithmic function. Draw the verticalasymptote with a dashed line.

73. f (x) = log2 (x − 5)74. f (x) = log2 x − 5

75. g (x) = log3 (x + 5) + 15

76. g (x) = log3 (x − 5) − 5

77. h (x) = log4 (−x) + 1

78. h (x) = 3 − log4 x

79. g (x) = ln (x − 2) + 3

80. g (x) = ln (x + 3) − 181. The population of a certain small town is growing according to the function

P (t) = 89,000(1.035) t , where t represents time in years since the last

ln ( 1e5 )

ln x =12



census. Use the function to estimate the population 8 12 years after the census

was taken.

82. The volume of sound L in decibels (dB) is given by the formula

L = 10 log (I/10−12), where I represents the intensity of the sound in

watts per square meter. Determine the volume of a sound with an intensity of0.5 watts per square meter.

PROPERTIES OF THE LOGARITHM

Evaluate without using a calculator.

83. log9 9

84. log8 1

85. log1/3 3

86. log ( 110 )

87. eln 17

88. 10 log 27

89. ln e63

90. log 1033

Expand completely.

91. log (100x 2)92. log5 (5x 3)

93.

94.

95.

log3 ( 3x 5

5 )ln ( 10

3x 2 )log2 ( 8x 2

y2z )



96.

97.

98.

Write as a single logarithm with coefficient 1.

99. log x + 2 log y − 3 log z

100. log2 5 − 3log2 x + 4log2 y

101. −2log5 x + log5 y − 5log5 (x − 1)

102. ln x − ln (x − 1) − ln (x + 1)

103. 3log2 x + 12 log2 y − 2

3 log2 z

104. 13 log x − 3 log y − 3

5 log z

105. log5 4 + 5log5 x − 13 (log5 y + 2log5 z)

106. ln x − 12 (ln y − 4 ln z)

SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS

Solve. Give the exact answer and the approximate answer rounded tothe nearest hundredth where appropriate.

107. 52x+1 = 125

108. 103x−2 = 100

109. 9x−3 = 81

110. 162x+3 = 8

111. 5x = 7

112. 32x = 5

log ( x 10

10y3z4 )ln (

3b a⎯⎯

√c4 )

log ( 20y3

x 2⎯ ⎯⎯⎯√3 )



113. 10x+2 − 3 = 7

114. e2x−1 + 2 = 3

115. 74x−1 − 2 = 9

116. 35x−2 + 5 = 7

117. 3 − e4x = 2

118. 5 + e3x = 4

119.

120.

Use the change of base formula to approximate the following to thenearest tenth.

121. log5 13

122. log2 27

123. log4 5

124. log9 0.81

125. log1/4 21

126. log2 5⎯⎯

√3

Solve.

127. log2 (3x − 5) = log2 (2x + 7)

128. ln (7x) = ln (x + 8)

129. log5 8 − 2log5 x = log5 2

130. log3 (x + 2) + log3 (x) = log3 8

131. log5 (2x − 1) = 2

132. 2log4 (3x − 2) = 4

41 + e5x

= 2100

1 + e3x=

12



133. 2 = log2 (x 2 − 4) − log2 3

134. log2 (x − 1) + log2 (x + 1) = 3

135. log2 x + log2 (x − 1) = 1

136. log4 (x + 5) + log4 (x + 11) = 2

137. log (2x + 5) − log (x − 1) = 1

138. ln x − ln (2x − 1) = 1

139. 2log2 (x + 4) = log2 (x + 2) + 3

140. 2log3 x = 1 + log3 (x + 6)141. log3 (x + 1) − 2log3 x = 1

142. log5 (2x) + log5 (x − 1) = 1

APPLICATIONS

Solve.

143. An amount of $3,250 is invested in an account that earns 4.6% annual interestthat is compounded monthly. Estimate the number of years for the amount inthe account to reach $4,000.

144. An amount of $2,500 is invested in an account that earns 5.5% annual interestthat is compounded continuously. Estimate the number of years for theamount in the account to reach $3,000.

145. How long does it take to double an investment made in an account that earns

6 34 % annual interest that is compounded continuously?

146. How long does it take to double an investment made in an account that earns

6 34 % annual interest that is compounded semi-annually?

147. In the year 2000 a certain small town had a population of 46,000 people. In theyear 2010 the population was estimated to have grown to 92,000 people. If thepopulation continues to grow exponentially at this rate, estimate thepopulation in the year 2016.



148. A fleet van was purchased new for $28,000 and 2 years later it was valued at$20,000. If the value of the van continues to decrease exponentially at this rate,determine its value 7 years after it is purchased new.

149. A website that has been in decline registered 4,200 unique visitors last monthand 3,600 unique visitors this month. If the number of unique visitorscontinues to decline exponentially, how many unique visitors would youexpect next month?

150. An initial population of 18 rabbits was introduced into a wildlife preserve. Thenumber of rabbits doubled in the first year. If the rabbit population continuesto grow exponentially at this rate, how many rabbits will be present 5 yearsafter they were introduced?

151. The half-life of sodium-24 is about 15 hours. How long will it take a50-milligram sample to decay to 10 milligrams?

152. The half-life of radium-226 is about 1,600 years. How long will it take an initialsample to decay to 30% of the original amount?

153. An archeologist discovered a bone tool artifact. After analysis, the artifact wasfound to contain 62% of the carbon-14 normally found in bone from the sameanimal. Given that carbon-14 has a half-life of 5,730 years, estimate the age ofthe artifact.

154. The half-life of radioactive iodine-131 is about 8 days. What percentage of aninitial sample accidentally released into the atmosphere do we expect toremain after 53 days?



ANSWERS

1. (f○g) (x) = 12x + 1 ; (g○f ) (x) = 12x − 9

3.

5.

7.

9. No, fails the HLT

11. Yes, passes the HLT

13. Proof

15. Proof

17. f −1 (x) = − 17 x + 3

7

19. g−1 (x) = x + 12⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

√

21. g−1 (x) = x+2x

23. h−1 (x) = − 1x−3

25. h−1 (x) = (x−2)3+72

27. f (−1) = 15 , f (0) = 1, f (3) = 125

29. g (−5) = 100,000 , g (0) = 1,g (2) = 1100

(f○g) (x) = 50x 2 + 5x − 2;(g○f ) (x) = 10x 2 + 5x − 10(f○g) (x) = 2 2x⎯ ⎯⎯⎯√ ;(g○f ) (x) = 8 x + 2⎯ ⎯⎯⎯⎯⎯⎯⎯⎯

√ − 2

(f○g) (x) = −x 2 − 7x + 9(x − 2)2

;

(g○f ) (x) = 1x 2 + 3x − 3



31.

33.

35.



37.

39. f (−3) ≈ 1.05 , f (−1) ≈ 1.37 , f ( 12 ) ≈ 2.65

41. p (−4) ≈ −271.99 , p (− 12 ) ≈ −7.24, p (0) = −4

43.

45.



47. $7,223.67

49. Approximately 1 gram

51. 2

53. −5

55. −2

57. 0

59. Undefined

61. 12

63. 8

65. −5

67. 125

69. 827

71. 0.001

73.



75.

77.

79.



81. 119,229 people

83. 1

85. −1

87. 17

89. 63

91. 2 + 2 log x

93. 1 + 5log3 x − log3 5

95. 3 + 2log2 x − 2log2 y − log2 z

97. ln 3 + ln b + 12 ln a − 4 ln c

99.

101.

103.

105.

107. 1

109. 5

111.

113. −1

115.

117. 0

119. 0

121. 1.6

log ( xy2

z3 )log5 ( y

x 2 (x − 1)5 )log2

x

3 y⎯⎯√

z2⎯ ⎯⎯⎯

√3

log5 ( 4x 5

yz2⎯ ⎯⎯⎯⎯⎯

√3 )

log (7)log (5) ≈ 1.21

log 7 + log 114 log 7

≈ 0.56



123. 1.2

125. −2.2

127. 12

129. 2

131. 13

133. ±4

135. 2

137. 158

139. 0

141.

143. 4.5 years

145. 10.27 years

147. About 139,446 people

149. 3,086 unique visitors

151. 35 hours

153. About 3,952 years old

1 + 13⎯ ⎯⎯⎯

√6



SAMPLE EXAM

1. Given f (x) = x 2 − x + 3 and g (x) = 3x − 1 find (f○g) (x) .2. Show that f (x) = 7x − 2

⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3 and g (x) = x 3+2

7 are inverses.

Find the inverse of the following functions:

3. f (x) = 12 x − 3

4. h (x) = x 2 + 3 where x ≥ 0

Sketch the graph.

5. f (x) = ex − 5

6. g (x) = 10−x

7. Joe invested $5,200 in an account earning 3.8% annual interest that iscompounded monthly. How much will be in the account at the end of 4 years?

8. Mary has $3,500 in a savings account earning 4 12 % annual interest that is

compounded continuously. How much will be in the account at the end of 3years?

Evaluate.

9. a. log3 81b. log2 ( 14 )c. log 1,000d. ln e

10. a. log4 2b. log9 ( 13 )c. ln e3

d. log1/5 25

Sketch the graph.

11. f (x) = log4 (x + 5) + 2

12. f (x) = − ln (x − 2)



13. Expand: log

100x 2y

z√

.

14. Write as a single logarithm with coefficient 1:

2log2 x + 13 log2 y − 3log2 z.

Evaluate. Round off to the nearest tenth.

15. a. log2 10b. ln 1c. log3 ( 15 )

Solve:

16. 23x−1 = 16

17. 37x+1 = 5

18. log5 (3x − 4) = log5 (2x + 7)

19. log3 (x 2 + 26) = 3

20. log2 x + log2 (2x + 7) = 2

21. log (2x + 3) = 1 + log (x + 1)22. Joe invested $5,200 in an account earning 3.8% annual interest that is

compounded monthly. How long will it take to accumulate a total of $6,200 inthe account?

23. Mary has $3,500 in a savings account earning 4 12 % annual interest that is

compounded continuously. How long will it take to double the amount in theaccount?

24. During the exponential growth phase, certain bacteria can grow at a rate of5.3% per hour. If 12,000 cells are initially present in a sample, construct anexponential growth model and use it to: a. Estimate the population of bacteriain 3.5 hours. b. Estimate the time it will take the population to double.

25. The half-life of caesium-137 is about 30 years. Approximate the time it will takea 20-milligram sample of caesium-137 to decay to 8 milligrams.



ANSWERS

1. (f○g) (x) = 9x 2 − 9x + 5

3. f −1 (x) = 2x + 6

5.

7. $6,052.18

9. a. 4b. −2c. 3d. 1

11.

13. 2 + 2 log x + log y − 12 log z

15. a. 3.3



b. 0c. −1.5

17.log 5−log 3

7 log 3

19. ±1

21. − 78

23. 15.4 years

25. 40 years



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This is “Exponential and Logarithmic Functions”, chapter 7 ...€¦ · Chapter 7 Exponential and Logarithmic Functions 7.1 Composition and Inverse Functions 1578 The horizontal